# Equivalent conditions for hereditarily Lindelof spaces

A topological space $X$ is Lindelof if every open cover $X$ has a countable subcollection that also is a cover of $X$. A topological space $X$ is hereditarily Lindelof if every subspace of $X$, with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

Theorem 1
Let $X$ be a topological space. The following conditions are equivalent.

1. The space $X$ is a hereditarily Lindelof space.
2. Every open subspace of $X$ is Lindelof.
3. For every uncountable subspace $Y$ of $X$, there exists a point $y \in Y$ such that every open subset of $X$ containing $y$ contains uncountably many points of $Y$.

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence $1 \longleftrightarrow 3$ is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

Proof of Theorem 1
The direction $1 \longrightarrow 2$ is immediate. The direction $2 \longrightarrow 3$ is straightforward.

$3 \longrightarrow 1$
We show $\text{not } 1 \longrightarrow \text{not } 3$. Suppose $T$ is a non-Lindelof subspace of $X$. Let $\mathcal{U}$ be an open cover of $T$ such that no countable subcollection of $\mathcal{U}$ can cover $T$. By a transfinite inductive process, choose a set of points $\left\{t_\alpha \in T: \alpha < \omega_1 \right\}$ and a collection of open sets $\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$ such that for each $\alpha < \omega_1$, $t_\alpha \in U_\alpha$ and $t_\alpha \notin \cup \left\{U_\beta: \beta<\alpha \right\}$. The inductive process is possible since no countable subcollection of $\mathcal{U}$ can cover $T$. Now let $Y=\left\{t_\alpha: \alpha<\omega_1 \right\}$. Note that each $U_\alpha$ can at most contain countably many points of $Y$, namely the points in $\left\{t_\beta: \beta \le \alpha \right\}$.

For each $\alpha$, let $V_\alpha$ be an open subset of $X$ such that $U_\alpha=V_\alpha \cap Y$. We can now conclude: for every point $t_\alpha$ of $Y$, there exists an open set $V_\alpha$ containing $t_\alpha$ such that $V_\alpha$ contains only countably many points of $Y$. This is the negation of condition 3. $\blacksquare$

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Remarks

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let $p \in X$. We say $p$ is a limit point of the set $Y \subset X$ if every open set containing $p$ contains a point of $Y$ different from $p$. Being a limit point of $Y$, we only know that each open set containing $p$ contain infinitely many points of $Y$ (assuming a $T_1$ space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if $p$ is a limit point of $Y$ satisfying condition 3, then $p$ is said to be a condensation point of $Y$. According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set $Y$ is a condensation point of $Y$.

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

Corollary 2
If the space $X$ is hereditarily Lindelof, then every uncountable subspace $Y$ of $X$ contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

Theorem 3
If the space $X$ is Lindelof, then every uncountable subspace $Y$ of $X$ has a limit point.

The condition “every uncountable subspace $Y$ of $X$ has a limit point” has another name. When a space \$latesatisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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$\copyright \ 2014 \text{ by Dan Ma}$

# An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space $C_p(X)$ that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space $X$ such that

• $C_p(X)$ is collectionwise normal and not paracompact,
• $C_p(X)$ is not Lindelof but contains a dense Lindelof subspace,
• $C_p(X)$ is not first countable but is a Frechet space,
• As a corollary of the previous point, $C_p(X)$ cannot contain a copy of the compact space $\omega_1+1$,
• $C_p(X)$ is homeomorphic to $C_p(X)^\omega$,
• $C_p(X)$ is not hereditarily normal,
• $C_p(X)$ is not metacompact.

A short and quick description of the space $X$ is that $X$ is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space $C_p(X)$ is intimately related to a $\Sigma$-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the $\Sigma$-products of copies of real lines. Deep results about $\Sigma$-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space $X=L_\tau$. In the discussion that follows, the Greek letter $\tau$ is always an uncountable cardinal number. Let $D_\tau$ be a set with cardinality $\tau$. Let $p$ be a point not in $D_\tau$. Let $L_\tau=D_\tau \cup \left\{p \right\}$. Consider the following topology on $L_\tau$:

• Each point in $D_\tau$ an isolated point, and
• open neighborhoods at the point $p$ are of the form $L_\tau-K$ where $K \subset D_\tau$ is countable.

It is clear that $L_\tau$ is a Lindelof space. The Lindelof space $L_\tau$ is sometimes called the one-point Lindelofication of the discrete space $D_\tau$ since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space $C_p(L_\tau)$. See this post for general information on the pointwise convergence topology of $C_p(Y)$ for any completely regular space $Y$.

All the facts about $C_p(X)=C_p(L_\tau)$ mentioned at the beginning follow from the fact that $C_p(L_\tau)$ is homeomorphic to the $\Sigma$-product of $\tau$ many copies of the real lines. Specifically, $C_p(L_\tau)$ is homeomorphic to the following subspace of the product space $\mathbb{R}^\tau$.

$\Sigma_{\alpha<\tau}\mathbb{R}=\left\{ x \in \mathbb{R}^\tau: x_\alpha \ne 0 \text{ for at most countably many } \alpha<\tau \right\}$

Thus understanding the function space $C_p(L_\tau)$ is a matter of understanding a $\Sigma$-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of $C_p(L_\tau)$ indicated above.

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The Homeomorphism

For each $f \in C_p(L_\tau)$, it is easily seen that there is a countable set $C \subset D_\tau$ such that $f(p)=f(y)$ for all $y \in D_\tau-C$. Let $W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}$. Then each $f \in W_0$ has non-zero values only on a countable subset of $D_\tau$. Naturally, $W_0$ and $\Sigma_{\alpha<\tau}\mathbb{R}$ are homeomorphic.

We claim that $C_p(L_\tau)$ is homeomorphic to $W_0 \times \mathbb{R}$. For each $f \in C_p(L_\tau)$, define $h(f)=(f-f(p),f(p))$. Here, $f-f(p)$ is the function $g \in C_p(L_\tau)$ such that $g(x)=f(x)-f(p)$ for all $x \in L_\tau$. Clearly $h(f)$ is well-defined and $h(f) \in W_0 \times \mathbb{R}$. It can be readily verified that $h$ is a one-to-one map from $C_p(L_\tau)$ onto $W_0 \times \mathbb{R}$. It is not difficult to verify that both $h$ and $h^{-1}$ are continuous.

We use the notation $X_1 \cong X_2$ to mean that the spaces $X_1$ and $X_2$ are homeomorphic. Then we have:

$C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R}) \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$

Thus $C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$. This completes the proof that $C_p(L_\tau)$ is topologically the $\Sigma$-product of $\tau$ many copies of the real lines.

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Looking at the $\Sigma$-Product

Understanding the function space $C_p(L_\tau)$ is now reduced to the problem of understanding a $\Sigma$-product of copies of the real lines. Most of the facts about $\Sigma$-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the $\Sigma$-product of separable metric spaces is collectionwise normal. Thus $C_p(L_\tau)$ is collectionwise normal. The $\Sigma$-product of spaces, each of which has at least two points, always contains a closed copy of $\omega_1$ with the ordered topology (see the lemma in this previous post). Thus $C_p(L_\tau)$ contains a closed copy of $\omega_1$ and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$:

$\sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}$

In this previous post, it is shown that $\sigma_\tau$ is a Lindelof space. Though $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is not Lindelof, it has a dense Lindelof subspace, namely $\sigma_\tau$.

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A space $Y$ is first countable if there exists a countable local base at each point $y \in Y$. A space $Y$ is a Frechet space (or is Frechet-Urysohn) if for each $y \in Y$, if $y \in \overline{A}$ where $A \subset Y$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $A$ such that the sequence converges to $y$. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number $\tau$, the product $\mathbb{R}^\tau$ is not first countable. In fact, any dense subspace of $\mathbb{R}^\tau$ is not first countable. In particular, the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$ is not first countable. In this previous post, it is shown that the $\Sigma$-product of first countable spaces is a Frechet space. Thus $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is a Frechet space.

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As a corollary of the previous point, $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space $\omega_1+1$ is not embeddable in $C_p(L_\tau)$. The interest in compact subspaces of $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is that any compact space that is topologically embeddable in a $\Sigma$-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

$\Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}$

Thus $C_p(L_\tau) \cong C_p(L_\tau)^\omega$. In particular, $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ due to the following observation:

$C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)$

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As a result of the peculiar fact that $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$, it can be concluded that $C_p(L_\tau)$, though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if $Y_1 \times Y_2$ is hereditarily normal, then either $Y_1$ is perfectly normal or every countably infinite subset of $Y_2$ is closed and discrete (see this previous post). The function space $C_p(L _\tau)$ is not perfectly normal since it contains a closed copy of $\omega_1$. On the other hand, there are plenty of countably infinite subsets of $C_p(L _\tau)$ that are not closed and discrete. As a Frechet space, $C_p(L _\tau)$ has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let $\left\{x_1,x_2,x_3,\cdots \right\}$ be an infinite subset of $D_\tau$ and consider the following:

$C=\left\{f_n: n=1,2,3,\cdots \right\}$

where $f_n$ is such that $f_n(x_n)=n$ and $f_n(x)=0$ for each $x \in L_\tau$ with $x \ne x_n$. Note that $C$ is not closed and not discrete since the points in $C$ converge to $g \in \overline{C}$ where $g$ is the zero-function. Thus $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since $C_p(L_\tau)$ is collectionwise normal and not paracompact, $C_p(L_\tau)$ can never be metacompact.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2014 \text{ by Dan Ma}$

# (Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $b \in \prod_{\alpha \in A} X_\alpha$, called the base point. The $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ is the following subspace of the product space $X$:

$\Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}$

In other words, the space $\Sigma_{\alpha \in A} X_\alpha$ is the subspace of the product space $X=\prod_{\alpha \in A} X_\alpha$ consisting of all points that deviate from the base point on at most countably many coordinates $\alpha \in A$. We also consider the following subspace of $\Sigma_{\alpha \in A} X_\alpha$.

$\sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}$

For convenience , we call $\Sigma_{\alpha \in A} X_\alpha$ the (upper case) Sigma-product (or $\Sigma$-product) of the spaces $X_\alpha$ and we call the space $\sigma$ the (lower case) sigma-product (or $\sigma$-product). Clearly, the space $\sigma$ is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor $X_\alpha$ is a separable metric space with at least two points, the $\Sigma$-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) $\Sigma$-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer $n=0,1,2,3,\cdots$, we define $\sigma_n$ as follows:

$\sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}$

Clearly, $\sigma=\bigcup_{n=0}^\infty \sigma_n$. We prove the following theorem. The fact that $\sigma$ is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$. We say $V$ is a standard basic open subset of the product space $X$ if $V$ is of the form $V=\prod_{\alpha \in A} V_\alpha$ such that each $V_\alpha$ is an open subset of the factor space $X_\alpha$ and $V_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. The finite set $F$ of all $\alpha \in A$ such that $V_\alpha \ne X_\alpha$ is called the support of the open set $V$.

Theorem 1
Let $\sigma$ be the $\sigma$-product of the separable metrizable spaces $\left\{X_\alpha: \alpha \in A \right\}$. For each $n$, let $\sigma_n$ be defined as above. The product space $\sigma_n \times Y$ is Lindelof for each non-negative integer $n$ and for all separable metric space $Y$.

Proof of Theorem 1
We prove by induction on $n$. Note that $\sigma_0=\left\{b \right\}$, the base point. Clearly $\sigma_0 \times Y$ is Lindelof for all separable metric space $Y$. Suppose the theorem hold for the integer $n$. We show that $\sigma_{n+1} \times Y$ for all separable metric space $Y$. To this end, let $\mathcal{U}$ be an open cover of $\sigma_{n+1} \times Y$ where $Y$ is a separable metric space. Without loss of generality, we assume that each element of $\mathcal{U}$ is of the form $V \times W$ where $V=\prod_{\alpha \in A} V_\alpha$ is a standard basic open subset of the product space $X=\prod_{\alpha \in A} X_\alpha$ and $W$ is an open subset of $Y$.

Let $\mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable subcollection of $\mathcal{U}$ such that $\mathcal{U}_0$ covers $\left\{b \right\} \times Y$. For each $j$, let $U_j=V_j \times W_j$ where $V_j=\prod_{\alpha \in A} V_{j,\alpha}$ is a standard basic open subset of the product space $X$ with $b \in V_j$ and $W_j$ is an open subset of $Y$. For each $j$, let $F_j$ be the support of $V_j$. Note that $\alpha \in F_j$ if and only if $V_{j,\alpha} \ne X_\alpha$. Also for each $\alpha \in F_j$, $b_\alpha \in V_{j,\alpha}$. Furthermore, for each $\alpha \in F_j$, let $V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}$. With all these notations in mind, we define the following open set for each $\beta \in F_j$:

$H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j$

Observe that for each point $y \in \sigma_{n+1}$ such that $y \in V^c_{j,\beta} \times T_\beta$, the point $y$ already deviates from the base point $b$ on one coordinate, namely $\beta$. Thus on the coordinates other than $\beta$, the point $y$ can only deviates from $b$ on at most $n$ many coordinates. Thus $\sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta)$ is homeomorphic to $V^c_{j,\beta} \times \sigma_n$. Note that $V^c_{j,\beta} \times W_j$ is a separable metric space. By inductive hypothesis, $V^c_{j,\beta} \times \sigma_n \times W_j$ is Lindelof. Thus there are countably many open sets in the open cover $\mathcal{U}$ that covers points of $H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$.

Note that

$\sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)$

To see that the left-side is a subset of the right-side, let $t=(x,y) \in \sigma_{n+1} \times Y$. If $t \in U_j$ for some $j$, we are done. Suppose $t \notin U_j$ for all $j$. Observe that $y \in W_j$ for some $j$. Since $t=(x,y) \notin U_j$, $x_\beta \notin V_{j,\beta}$ for some $\beta \in F_j$. Then $t=(x,y) \in H_{j,\beta}$. It is now clear that $t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$. Thus the above set equality is established. Thus one part of $\sigma_{n+1} \times Y$ is covered by countably many open sets in $\mathcal{U}$ while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of $\mathcal{U}$ covers $\sigma_{n+1} \times Y$. $\blacksquare$

Corollary 2
It follows from Theorem 1 that

• If each factor space $X_\alpha$ is a separable metric space, then each $\sigma_n$ is a Lindelof space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a Lindelof space.
• If each factor space $X_\alpha$ is a compact separable metric space, then each $\sigma_n$ is a compact space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that $\Sigma$-product of compact spaces is countably compact. Thus $\Sigma_{\alpha \in A} X_\alpha$ is a countably compact space if each $X_\alpha$ is compact. Note that each $\sigma_n$ is a closed subset of $\Sigma_{\alpha \in A} X_\alpha$ and is thus countably compact. Being a Lindelof space, each $\sigma_n$ is compact. It follows that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space. $\blacksquare$

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each $\alpha \in A$, let $X_\alpha$ be a separable metric space with at least two points. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is collectionwise normal (see this previous post). According to the lemma in this previous post, the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ contains a closed copy of $\omega_1$. Thus the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is not Lindelof. It is clear that the $\sigma$-product is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. By Corollary 2, the $\sigma$-product is a Lindelof subspace of $\Sigma_{\alpha \in A} X_\alpha$.

Using specific factor spaces, if each $X_\alpha=\mathbb{R}$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each $X_\alpha=[0,1]$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense $\sigma$-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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$\copyright \ 2014 \text{ by Dan Ma}$

# Cp(X) where X is a separable metric space

Let $\tau$ be an uncountable cardinal. Let $\prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau}$ be the Cartesian product of $\tau$ many copies of the real line. This product space is not normal since it contains $\prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1}$ as a closed subspace. However, there are dense subspaces of $\mathbb{R}^{\tau}$ are normal. For example, the $\Sigma$-product of $\tau$ copies of the real line is normal, i.e., the subspace of $\mathbb{R}^{\tau}$ consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of $\mathbb{R}^{\tau}$ that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space $C_p(X)$ where $X$ is a separable metrizable space.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

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$C_p(X)$ when $X$ is a separable metric space

In the remainder of the post, $X$ denotes a separable metrizable space. Then, $C_p(X)$ is more than normal. The function space $C_p(X)$ has the following properties:

• normal,
• Lindelof (hence paracompact and collectionwise normal),
• hereditarily Lindelof (hence hereditarily normal),
• hereditarily separable,
• perfectly normal.

All such properties stem from the fact that $C_p(X)$ has a countable network whenever $X$ is a separable metrizable space.

Let $L$ be a topological space. A collection $\mathcal{N}$ of subsets of $L$ is said to be a network for $L$ if for each $x \in L$ and for each open $O \subset L$ with $x \in O$, there exists some $A \in \mathcal{N}$ such that $x \in A \subset O$. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for $C_p(X)$, let $\mathcal{B}$ be a countable base for the domain space $X$. For each $B \subset \mathcal{B}$ and for any open interval $(a,b)$ in the real line with rational endpoints, consider the following set:

$[B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}$

There are only countably many sets of the form $[B,(a,b)]$. Let $\mathcal{N}$ be the collection of sets, each of which is the intersection of finitely many sets of the form $[B,(a,b)]$. Then $\mathcal{N}$ is a network for the function space $C_p(X)$. To see this, let $f \in O$ where $O=\bigcap_{x \in F} [x,O_x]$ is a basic open set in $C_p(X)$ where $F \subset X$ is finite and each $O_x$ is an open interval with rational endpoints. For each point $x \in F$, choose $B_x \in \mathcal{B}$ with $x \in B_x$ such that $f(B_x) \subset O_x$. Clearly $f \in \bigcap_{x \in F} \ [B_x,O_x]$. It follows that $\bigcap_{x \in F} \ [B_x,O_x] \subset O$.

Examples include $C_p(\mathbb{R})$, $C_p([0,1])$ and $C_p(\mathbb{R}^\omega)$. All three can be considered subspaces of the product space $\mathbb{R}^c$ where $c$ is the cardinality of the continuum. This is true for any separable metrizable $X$. Note that any separable metrizable $X$ can be embedded in the product space $\mathbb{R}^\omega$. The product space $\mathbb{R}^\omega$ has cardinality $c$. Thus the cardinality of any separable metrizable space $X$ is at most continuum. So $C_p(X)$ is the subspace of a product space of $\le$ continuum many copies of the real lines, hence can be regarded as a subspace of $\mathbb{R}^c$.

A space $L$ has countable extent if every closed and discrete subset of $L$ is countable. The $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ of the separable metric spaces $\left\{X_\alpha: \alpha \in A \right\}$ is a dense and normal subspace of the product space $\prod_{\alpha \in A} X_\alpha$. The normal space $\Sigma_{\alpha \in A} X_\alpha$ has countable extent (hence collectionwise normal). The examples of $C_p(X)$ discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Weakly Lindelof spaces

The weakly Lindelof property is a natural weakening of the familiar Lindelof property. In this post, we discuss some of the basic properties of weakly Lindelof spaces.

We consider topological spaces that are at least $T_1$ (i.e. finite sets are closed) and regular. A space $X$ is said to be Lindelof if for any open cover $\mathcal{U}$ of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $X=\bigcup \mathcal{V}$. A natural weakening of the Lindelof property is that we only require the countable $\mathcal{V}$ to cover a dense subset of the space $X$. Specifically, a space $X$ is said to be a weakly Lindelof space if for any open cover $\mathcal{U}$ of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$.

The notion of weakly Lindelof has a brief mention in the Encyclopedia of General Topology (see page 183 in [4]), pointing out a connection to Banach space theory. Furthermore, assuming CH, the weakly Lindelof subspaces of $\beta \mathbb{N}$ are precisely those subspaces which are $C^*$-embedded into $\beta \mathbb{N}$. In this post, we focus on the basic properties.

Clearly separable spaces and Lindelof spaces are weakly Lindelof. Another obvious property that implies weakly Lindelof is the existence of a dense Lindelof subspace. It is slightly less obvious that the countable chain condition implies the weakly Lindelof property. We have the following implications.

All the affirmative implications in the above diagram cannot be reversed (see Examples 1, 2 and 3 below).

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Some Cardinal Functions

Some of the properties discussed below can be described by cardinal functions, e.g., Lindelof number and weak Lindelof numbers. So we describe these before going into the basic properties. Let $X$ be a space. The Lindelof number of the space $X$, denoted by $L(X)$, is the least cardinal number $\mathcal{K}$ such that every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V} \subset \mathcal{U}$ with $\lvert \mathcal{V} \lvert \le \mathcal{K}$ such that $\mathcal{V}$ is a cover of $X$. When $L(X)=\omega$, we say that the space is Lindelof.

The weak Lindelof number of the space $X$, denoted by $wL(X)$, is the least cardinal number $\mathcal{K}$ such that every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V} \subset \mathcal{U}$ with $\lvert \mathcal{V} \lvert \le \mathcal{K}$ such that $X=\overline{\bigcup \mathcal{V}}$. When $wL(X)=\omega$, we say that the space is weakly Lindelof.

The character at $x \in X$, denoted by $\chi(x,X)$, is the least cardinal number of a local base at the point $x \in X$. The character of the space $X$, denoted by $\chi(X)$, is the supremum of all the cardinal numbers $\chi(x,X)$ over all $x \in X$. When $\chi(X)=\omega$, we say that $X$ is first countable.

The cellularity of the space $X$, denoted by $c(X)$, is the least infinite cardinal number $\mathcal{K}$ such that every collection of pairwise disjoint non-empty open subsets of $X$ has cardinality $\le \mathcal{K}$. When $c(X)=\omega$, we say that $X$ has the countable chain condition.

The extent of the space $X$, denoted by $e(X)$, is the least infinite cardinal number $\mathcal{K}$ such that if $A$ is a closed and discrete subset of $X$, then $\lvert A \lvert \le \mathcal{K}$. If $e(X)=\omega$, then $X$ is said to have countable extent (there are no uncountable closed and discrete subset). It is well known that Lindelof spaces have countable extent. The Lindelof number and the extent is related by the inequality: $e(X) \le L(X)$.

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Basic Properties

Weakly Lindelof spaces behave differently from Lindelof spaces in some ways. For example, closed subsets of a weakly Lindelof space do not have to be weakly Lindelof. In other ways, weakly Lindelof spaces and Lindelof spaces behave similarly. For example, the product of weakly Lindelof spaces needs not be weakly Lindelof and that every continuous image of a weakly Lindelof space is weakly Lindelof. Any Lindelof, Hausdorff and first countable space has cardinality no more than continuum. There is a similar theorem for weakly Lindelof spaces. Despite all these similarities with Lindelof spaces, the weak Lindelof property is a very weak property. It is well known that every Lindelof space has countable extent. There is no bound on the extent of weakly Lindelof spaces. The extent of a weakly Lindelof space can be arbitrarily large (see Example 4 below).

We discuss the following properties of weakly Lindelof spaces.

1. Any space with the countable chain condition is weakly Lindelof.
2. Any paracompact weakly Lindelof space is Lindelof.
3. Every continuous image of a weakly Lindelof space is weakly Lindelof.
4. The product of a compact space and a weakly lindelof space is weakly Lindelof.
5. The product of two Lindelof spaces needs not be weakly Lindelof.
6. Any normal first countable weakly Lindelof space has cardinality $\le 2^\omega$.
7. For any infinite cardinal $\mathcal{K}$, there exists a weakly Lindelof space $X$ such that $e(X) \ge \mathcal{K}$, i.e., the extent is at least $\mathcal{K}$. See Example 4 below.

Proof of 1
A space $X$ has the countable chain condition (has the CCC or is CCC for short) if there exists no uncountable collection of non-empty pairwise disjoint open subsets of $X$. “CCC $\Longrightarrow$ weakly Lindelof” follows from the following theorem (proved in this previous post).

Theorem
A space $X$ has the CCC if and only if for every collection $\mathcal{U}$ of non-empty open subsets of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$.

To finish off, let $\mathcal{U}$ be an open cover of $X$. By the theorem, there exists a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$. This means that $X=\overline{\bigcup \mathcal{V}}$. $\blacksquare$

Even though CCC implies weakly Lindelof, CCC does not imply the stronger property of having a dense Lindelof subspace (see Example 3 below).

The proof of 1 can be generalized to show that $wL(X) \le c(X)$ for any space $X$. However, the inequality cannot be made an equality. In fact, the inequality $wL(X) \le c(X)$ can be made as wide as one wishes. Specifically, we can keep $wL(X)=\omega$ while making $c(X)$ as large as one wishes (see Example 2 below). Thus the notions of countable chain condition and the weakly Lindelof property are far apart.

Proof of 2
Let $\mathcal{U}$ be an open cover of a paracompact weakly Lindelof space $X$. Using the regularity of the space, there is an open refinement $\mathcal{V}$ of $\mathcal{U}$ for each $V \in \mathcal{V}$, $\overline{V} \subset U$ for some $U \in \mathcal{U}$. Using the paracompactness, let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$. Using the weakly Lindelof property, choose a countable $\mathcal{C} \subset \mathcal{W}$ such that $X=\overline{\bigcup \mathcal{C}}$. With the collection $\mathcal{C}$ being locally finite, we have $X=\overline{\bigcup \mathcal{C}}=\bigcup \left\{\overline{C}: C \in \mathcal{C} \right\}$. Thus every point of $X$ belongs to some $\overline{C}$ for some $C \in \mathcal{C}$. Tracing from $\mathcal{C}$ to $\mathcal{W}$, to $\mathcal{V}$ and then to $\mathcal{U}$, we see that for every $C \in \mathcal{C}$, $\overline{C} \subset U$ for some $U \in \mathcal{U}$. It follows that a countable subcollection of $\mathcal{U}$ is a cover of $X$. This completes the proof of bullet point 2.

This result implies that in any metrizable space, the weakly Lindelof number coincides with the Lindelof number. So in metrizable spaces, the weak Lindelof number is just as good as an indicator of weight as the other cardinal functions such as density and Lindelof number.

Among CCC spaces, paracompactness and the Lindelof property coincide. This result shows that among weakly Lindelof spaces, paracompactness and the Lindelof property also coincide. $\blacksquare$

The proof of 3 is straightforward. It is very similar to the proof that continuous image of a Lindelof space is Lindelof.

Proof of 4
The proof that the product of a compact space and a weakly Lindelof space is weakly Lindelof makes use of the tube lemma, as in the proof that the product of a compact space and a Lindelof space is Lindelof.

Let $X$ be weakly Lindelof. Let $Y$ be compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\left\{ x \right\} \times Y$. By the tube lemma, there exists some open set $O_x \subset X$ such that $\left\{ x \right\} \times Y \subset O_x \times Y \subset \bigcup \mathcal{U}_x$.

Since $X$ is weakly Lindelof, there exists a countable $A \subset X$ such that $X=\overline{\bigcup \limits_{x \in A} O_x}$. Let $\mathcal{U}_A=\bigcup \limits_{x \in A} \mathcal{U}_x$. It is clear that $\mathcal{U}_A$ is a countable subcollection of $\mathcal{U}$. Note that the set $\bigcup \limits_{x \in A} (O_x \times Y)$ is dense in $X \times Y$. Thus the set $\bigcup \bigcup \limits_{x \in A} \mathcal{U}_x$ is dense in $X \times Y$ too. Thus $X \times Y=\overline{\bigcup \bigcup \limits_{x \in A} \mathcal{U}_x}$. This completes the proof that $X \times Y$ is weakly Lindelof. $\blacksquare$

Proof of 5
An example of two Lindelof spaces whose product is not weakly Lindelof is provided in [3]. $\blacksquare$

Discussion of 6
Any Lindelof first countable Hausdorff space has cardinality no more than continuum (discussed in this previous post). This fact is a specific case of the general theorem that

$\lvert X \lvert \le 2^{\chi(X) \cdot L(X)}$

for any Hausdorff space $X$. Hence, the cardinality of any first countable Lindelof space is bounded by $2^\omega$. It is interesting that there is an analogous result for weakly Lindelof space. In [2], the following inequality was proved:

$\lvert X \lvert \le 2^{\chi(X) \cdot wL(X)}$

for any normal space (Theorem 2.1 in [2]). Thus the cardinality of any normal weakly Lindelof space is bounded by $2^\omega$.

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Examples

Example 1 and Example 2 below use Lindelof or compact spaces that do not have the CCC as starting point. Here’s several examples of Lindelof non-CCC spaces:

• One-point Lindelofication of an uncountable set. The space is denoted by $L(\mathcal{K})$ and is the set $\left\{p \right\} \cup D(\mathcal{K})$ where $D(\mathcal{K})$ is the discrete space of cardinality $\mathcal{K}$ and $p$ is a point not in $D(\mathcal{K})$. The open neighborhoods at $p$ have the form $\left\{p \right\} \cup (D(\mathcal{K})-C)$ where $C \subset D(\mathcal{K})$ is countable.
• The space $\omega_1+1$ with the order topology. Note that $\omega_1+1$ is the immediate successor of $\omega_1$, the first uncountable ordinal. See here.
• The unit square $[0,1] \times [0,1]$ with the lexicographic order. See here.
• The Alexandroff Double Circle. See here.

In the above four spaces, the first one is Lindelof and the other three are compact. All four do not have the countable chain condition.

Example 1
A non-Lindelof space $X_1$ that has a dense Lindelof subspace. As a bonus, this space does not have the CCC.

The idea is to start with a space that has a countable dense set of isolated points and an uncountable closed and discrete subset. One such space is a so called psi-space, a space defined using an uncountable almost disjoint family of subsets of $\omega$. Then replace each of the countably many isolated points with a copy of one of the above examples of a Lindelof space without the CCC.

Let $\omega$ the first infinite ordinal (or the set of all nonnegative integers). Let $\mathcal{A}$ be an uncountable almost disjoint family of subsets of $\omega$ (for the purpose of this example, it does not have to be an maximal almost disjoint family). Let $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$, where each $n \in \omega$ is isolated and each $A \in \mathcal{A}$ has open neighborhoods of the form $\left\{A \right\} \cup (A-F)$ where $F \subset \omega$ is finite. For a more detailed discussion about Psi-space, see this previous post.

Let $Y$ be any one of the above Lindelof space that is not CCC. For each $n \in \omega$, let $Y_n=Y \times \left\{n \right\}$. So the $Y_n$ are distinct copies of the space $Y$. The underlying set of this example is the following set:

$X_1=\mathcal{A} \cup \bigcup \limits_{n \in \omega} Y_n$

The topology on $X_1$ is defined in such a way that each $Y_n$ is considered a copy of the space $Y$ and each $A \in \mathcal{A}$ has open neighborhoods of the form:

$\left\{A \right\} \cup \bigcup \limits_{n \in A-F} Y_n$

where $F \subset \omega$ is finite. The union of all $Y_n$ is a dense Lindelof subspace of $X_1$. The set $\mathcal{A}$ is an uncountable closed and discrete subset of $X_1$. Thus $X_1$ is not Lindelof. Each $Y_n$ has uncountably many disjoint open sets. Thus $X_1$ does not have the CCC. This example shows that the existence of a dense Lindelof subspace implies neither the CCC nor the Lindelof property.

Example 2
A weakly Lindelof non-CCC space $X_2$.

Let $X$ be any one of the above three non-CCC compact spaces. Let $Y$ be any space with the CCC, hence is weakly Lindelof. Let $X_2=X \times Y$. Then $X \times Y$ is weakly Lindelof. It is also clear that $X \times Y$ does not have the CCC. This example shows that the weakly Lindelof property does not imply the countable chain condition.

This example shows that $\omega=wL(X_2). In fact, it is possible to make $c(X_2)$ as large as possible. In the definition of $X \times Y$ in this example, let $X$ be the one-point Lindelofication $L(\mathcal{K})$ and $Y$ be any CCC space. Then $c(L(\mathcal{K}))$ can be made as large as possible. Hence $c(X \times Y)$ can be made as large as possible.

Example 3
A CCC space $X_3$ that has no dense Lindelof subspace.

This example is found in a paper of Arhangel’skii (Theorem 1.1 in [1]). Let $C(\omega_1+1)$ be the set of all continuous real-valued functions defined on $\omega_1+1$. The set $C(\omega_1+1)$ endowed with the pointwise convergence topology is typically denoted by $C_p(\omega_1+1)$. The space we want to use is $X_3=C_p(\omega_1+1)$.

The space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$. Thus $C_p(\omega_1+1)$ has the CCC. In [1], it is shown that $C_p(\omega_1+1)$ does not contain a dense normal subspace. Hence it does not contain a dense Lindelof subspace. The proof that $C_p(\omega_1+1)$ does not contain a dense normal subspace is a deep and non-trivial result.

The example $X_3=C_p(\omega_1+1)$ shows that even though CCC implies the weakly Lindelof property, it cannot give the stronger property of the existence of a dense Lindelof subspace. It is also an example showing that the implication “existence of a dense Lindelof subspace $\Longrightarrow$ weakly Lindelof” cannot be reversed.

Example 4
An weakly Lindelof space $X_4$ such that the extent can be made arbitrarily large.

Let $\mathcal{K}$ be any uncountable cardinal. Let $W$ be a discrete space of cardinality $\mathcal{K}$. Let $\beta W$ be the Stone-Cech compactification of $W$. Consider the ordinal $S=\omega+1$ with the order topology (can just think of it as a sequence of isolated points converging to the limit $\omega$). The space $X_4$ is defined as follows:

$X_4=\beta W \times S-(\beta W-W) \times \left\{\omega \right\}$

Note that $\beta W \times \omega$ is a $\sigma$-compact dense subspace of $X_4$. Hence $X_4$ is weakly Lindelof. On the other hand, the set $W \times \left\{\omega \right\}$ is a closed and discrete subset of $X_4$. Since the cardinality of $W$ can be made arbitrarily large, the extent of $X_4$ can be made arbitrarily large. Thus there is no upper bound on the extent of weakly Lindelof spaces (unlike Lindelof spaces).

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Reference

1. Arhangel’skii A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Bell M., Ginsburg J., Woods G., Cardinal Inequalities for Topological Spaces Involving the Weak Lindelof Number, Pacific J. Math., 79, no. 1, 37-45, 1978.
3. Hajnal A., Juhasz I., On the Products of Weakly Lindelof Spaces, Proc. Amer. Math. Soc., 130, no. 1, 454-456, 1975.
4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A theorem about CCC spaces

It is a well known result in general topology that in any regular space with the countable chain condition, paracompactness and the Lindelof property are equivalent. The proof of this result hinges on one theorem about the spaces with the countable chain condition. In this post we are to put the spotlight on this theorem (Theorem 1 below) and then use it to prove a few results. These results indicate that in a space with the countable chain condition with some weaker covering property is either Lindelof or paracompact.

This post is centered on a theorem about the CCC property (Theorem 1 and Theorem 1a below). So it can be considered as a continuation of a previous post on CCC called Some basic properties of spaces with countable chain condition. The results that are derived from Theorem 1 are also found in [2]. But the theorem concerning CCC is only a small part of that paper among several other focuses. In this post, the exposition is to explain several interesting theorems that are derived from Theorem 1. One of the theorems is the statement that every locally compact metacompact perfectly normal space is paracompact, a theorem originally proved by Arhangelskii (see Theorem 11 below).

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CCC Spaces

All spaces under consideration are at least $T_1$ and regular. A space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if $X$ has the CCC, we also say that $X$ is a CCC space or $X$ is CCC.

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The theorem of CCC spaces we want to discuss has to do with collections of open sets that are “nice”. We first define what we mean by nice. Let $\mathcal{A}$ be a collection of non-empty subsets of the space $X$. The collection $\mathcal{A}$ is said to be point-finite (point-countable) if each point of $X$ belongs to only finitely (countably) many sets in $\mathcal{A}$.

Now we define what we mean by “nice” collection of open sets. The collection $\mathcal{A}$ is said to be locally finite (locally countable) at a point $x \in X$ if there exists an open set $O \subset X$ with $x \in O$ such that $O$ meets at most finitely (countably) many sets in $\mathcal{A}$. The collection $\mathcal{A}$ is said to be locally finite (locally countable) if it is locally finite (locally countable) at each $x \in X$.

The property of being a separable space implies the CCC. The reverse is not true. However the CCC property is still a very strong property. The CCC property is equivalent to the property that if a collection of non-empty open sets is “nice” on a dense set of points, then the collection of open sets is a countable collection. The following is a precise statement.

Theorem 1

Let $X$ be a CCC space. Then if $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The collections of open sets in the above theorem do not have to be open covers. However, if they are open covers, the theorem can tie CCC spaces with some covering properties. As long as the space has the CCC, any open cover that is locally-countable on a dense set must be countable. Looking at it in the contrapositive angle, in a CCC space, any uncountable open cover is not locally-countable in some open set.

Proof of Theorem 1
Let $\mathcal{U}$ be a collection of open subsets of $X$ such that the set $D(\mathcal{U})$ as defined above is dense in the open subspace $\bigcup \mathcal{U}$. We show that $\mathcal{U}$ is countable. Suppose not.

For each $U \in \mathcal{U}$, since $U \cap D(\mathcal{U}) \ne \varnothing$, we can choose a non-empty open set $f(U) \subset U$ such that $f(U)$ has non-empty intersection with only countably many sets in $\mathcal{U}$. Let $\mathcal{U}_f$ be the following collection:

$\mathcal{U}_f=\left\{f(U): U \in \mathcal{U} \right\}$

For $H,K \in \mathcal{U}_f$, by a chain from $H$ to $K$, we mean a finite collection

$\left\{W_1,W_2,\cdots,W_n \right\} \subset \mathcal{U}_f$

such that $H=W_1$, $K=W_n$ and $W_j \cap W_{j+1} \ne \varnothing$ for any $1 \le j . For each open set $W \in \mathcal{U}_f$, define $\mathcal{C}(W)$ and $\mathcal{E}(W)$ as follows:

$\mathcal{C}(W)=\left\{V \in \mathcal{U}_f: \text{there exists a chain from } W \text{ to } V \right\}$

$\mathcal{E}(W)=\bigcup \mathcal{C}(W)$

One observation we make is that for $W_1,W_2 \in \mathcal{U}_f$, if $\mathcal{E}(W_1) \cap \mathcal{E}(W_2) \ne \varnothing$, then $\mathcal{C}(W_1)=\mathcal{C}(W_2)$ and $\mathcal{E}(W_1)=\mathcal{E}(W_2)$. So the distinct $\mathcal{E}(W)$ are pairwise disjoint. Because the space $X$ has the CCC, there can be only countably many distinct open sets $\mathcal{E}(W)$. Thus there can be only countably many distinct collections $\mathcal{C}(W)$.

Note that each $\mathcal{C}(W)$ is a countable collection of open sets. Each $V \in \mathcal{U}_f$ meets only countably many open sets in $\mathcal{U}$. So each $V \in \mathcal{U}_f$ can meet only countably many sets in $\mathcal{U}_f$, since for each $V \in \mathcal{U}_f$, $V \subset U$ for some $U \in \mathcal{U}$. Thus for each $W \in \mathcal{U}_f$, in considering all finite-length chain starting from $W$, there can be only countably many open sets in $\mathcal{U}_f$ that can be linked to $W$. Thus $\mathcal{C}(W)$ must be countable. In taking the union of all $\mathcal{C}(W)$, we get back the collection $\mathcal{U}_f$. Thus we have:

$\mathcal{U}_f=\bigcup \limits_{W \in \mathcal{U}_f} \mathcal{C}(W)$

Because the space $X$ is CCC, there are only countably many distinct collections $\mathcal{C}(W)$ in the above union. Each $\mathcal{C}(W)$ is countable. So $\mathcal{U}_f$ is a countable collection of open sets.

Furthermore, each $U \in \mathcal{U}$ contains at least one set in $\mathcal{U}_f$. From the way we choose sets in $\mathcal{U}_f$, we see that for each $V \in \mathcal{U}_f$, $V=f(U) \subset U$ for at most countably many $U \in \mathcal{U}$. The argument indicates that we have a one-to-countable mapping from $\mathcal{U}_f$ to $\mathcal{U}$. Thus the original collection $\mathcal{U}$ must be countable. $\blacksquare$

The property in Theorem 1 is actually equivalent to the CCC property. Just that the proof of Theorem 1 represents the hard direction that needs to be proved. Theorem 1 can be expanded to be the following theorem.

Theorem 1a

Let $X$ be a space. Then the following conditions are equivalent.

1. The space $X$ has the CCC.
2. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

3. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that $\mathcal{U}$ is locally-countable at every point in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The direction $1 \rightarrow 2$ has been proved above. The directions $2 \rightarrow 3$ and $3 \rightarrow 1$ are straightforward.

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Tying Theorem 1 to “Nice” Open Covers

One easy application of Theorem 1 is to tie it to locally-finite and locally-countable open covers. We have the following theorem.

Theorem 2

In any CCC space, any locally-countable open cover must be countable. Thus any locally-finite open cover must also be countable.

Theorem 2 gives the well known result that any CCC paracompact space is Lindelof (see Theorem 5 below). In fact, Theorem 2 gives the result that any CCC para-Lindelof space is Lindelof (see Theorem 6 below). A space $X$ is para-Lindelof if every open cover has a locally-countable open refinement.

Can Theorem 2 hold for point-finite covers (or point-countable covers)? The answer is no (see Example 1 below). With the additional property of having a Baire space, we have the following theorem.

Theorem 3

In any Baire space with the CCC, any point-finite open cover must be countable.

A Space $X$ is a Baire space if $U_1,U_2,U_3,\cdots$ are dense open subsets of $X$, then $\bigcap \limits_{j=1}^\infty U_j \ne \varnothing$. For more information about Baire spaces, see this previous post.
.

Proof of Theorem 3
Let $X$ be a Baire space with the CCC. Let $\mathcal{U}$ be a point-finite open cover of $X$. Suppose that $\mathcal{U}$ is uncountable. We show that this assumption with lead to a contradiction. Thus $\mathcal{U}$ must be countable.

By Theorem 1, there exists an open set $V \subset X$ such that $\mathcal{U}$ is not locally-countable at any point in $V$. For each positive integer $n$, let $H_n$ be the following:

$H_n=\left\{x \in V: x \text{ is in at most } n \text{ sets in } \mathcal{U} \right\}$

Note that $V=\bigcup \limits_{j=1}^\infty H_j$. Furthermore, each $H_n$ is a closed set in the space $V$. Since $X$ is a Baire space, every non-empty open subset of $X$ is of second category (i.e. it cannot be a union of countably many closed and nowhere dense sets). Thus it cannot be that each $H_n$ is nowhere dense in $V$. For some $n$, $H_n$ is not nowhere dense. There must exist some open $W \subset V$ such that $H_n \cap W$ is dense in $W$. Because $H_n$ is closed, $W \subset H_n$.

Choose $y \in W$. The point $y$ is in at most $n$ open sets in $\mathcal{U}$. Let $U_1,U_2,\cdots,U_m \in \mathcal{U}$ such that $y \in \bigcap \limits_{j=1}^m U_j$. Clearly $1 \le m \le n$. Let $U=W \cap U_1 \cap \cdots \cap U_m$. Note that $y \in U \subset H_n \subset V$.

Every point in $U$ belongs to at most $n$ many sets in $\mathcal{U}$ and already belong to $m$ sets in $\mathcal{U}$. So each point in $U$ can belong to at most $n-m$ additional open sets in $\mathcal{U}$. Consider the case $n-m=0$ and the case $n-m>0$. We show that each case leads to a contradiction.

Suppose that $n-m=0$. Then each point of $U$ can only meet $n$ open sets in $\mathcal{U}$, namely $U_1,U_2,\cdots,U_m$. This contradicts that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Suppose that $k=n-m>0$. Let $\mathcal{U}^*=\mathcal{U}-\left\{U_1,\cdots,U_m \right\}$. Let $\mathcal{M}$ be the following collection:

$\mathcal{M}=\left\{U \cap \bigcap \limits_{O \in M} O \ne \varnothing: M \subset \mathcal{U}^* \text{ and } \lvert M \lvert=k \right\}$

Each element of $\mathcal{M}$ is an open subset of $U$ that is the intersection of exactly $n$ many open sets in $\mathcal{U}$. So $\mathcal{M}$ is a collection of pairwise disjoint open sets. The open set $U$ as a topological space has the CCC. So $\mathcal{M}$ is at most countable. Thus the open set $U$ meets at most countably many open sets in $\mathcal{U}$, contradicting that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Both cases $n-m=0$ and $n-m>0$ lead to contradiction. So $\mathcal{U}$ must be countable. The proof to Theorem 3 is completed. $\blacksquare$

As a corollary to Theorem 3, we have the result that every Baire CCC metacompact space is Lindelof.

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Some Applications of Theorems 2 and 3

In proving paracompactness in some of the theorems, we need a theorem involving the concept of star-countable open cover. A collection $\mathcal{A}$ of subsets of a space $X$ is said to be star-finite (star-countable) if for each $A \in \mathcal{A}$, only finitely (countably) many sets in $\mathcal{A}$ meets $A$, i.e., the following set

$\left\{B \in \mathcal{A}: B \cap A \ne \varnothing \right\}$

is finite (countable). The proof of the following theorem can be found in Engleking (see the direction (iv) implies (i) in the proof of Theorem 5.3.10 on page 326 in [1]).

Theorem 4

If every open cover of a regular space $X$ has a star-countable open refinement, then $X$ is paracompact.

As indicated in the above section, Theorem 2 and Theorem 3 have some obvious applications. We have the following theorems.

Theorem 5

Let $X$ be a CCC space. Then $X$ is paracompact if and only of $X$ is Lindelof.

Proof of Theorem 5
The direction $\Longleftarrow$ follows from the fact that any regular Lindelof space is paracompact.

The direction $\Longrightarrow$ follows from Theorem 2. $\blacksquare$

Theorem 6

Every CCC para-Lindelof space is Lindelof.

Proof of Theorem 6
This also follows from Theorem 2. $\blacksquare$

Theorem 7

Every Baire CCC metacompact space is Lindelof.

Proof of Theorem 7
Let $X$ be a Baire CCC metacompact space. Let $\mathcal{U}$ be an open cover of $X$. By metacompactness, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. By Theorem 3, $\mathcal{V}$ must be countable. $\blacksquare$

Theorem 8

Every Baire CCC hereditarily metacompact space is hereditarily Lindelof.

Proof of Theorem 8
Let $X$ be a Baire CCC hereditarily metacompact space. To show that $X$ is hereditarily Lindelof, it suffices to show that every non-empty open subset is Lindelof. Let $Y \subset X$ be open. Then $Y$ has the CCC and is also metacompact. Being a Baire space is hereditary with respect to open subspaces. So $Y$ is a Baire space too. By Theorem 7, $Y$ is Lindelof. $\blacksquare$

Theorem 9

Every locally CCC regular para-Lindelof space is paracompact.

Proof of Theorem 9
A space is locally CCC if every point has an open neighborhood that has the CCC. Let $X$ be a regular space that is locally CCC and para-Lindelof. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the para-Lindelof assumption, let $\mathcal{V}$ be a locally-countable open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a locally-countable open cover. By Theorem 2, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 10

Every locally CCC regular metacompact Baire space is paracompact.

Proof of Theorem 10
Let $X$ be a regular space that is locally CCC and is a metacompact Baire space. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the metacompact assumption, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too. Each open set in $\mathcal{V}$ is also a Baire space.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a point-finite open cover. By Theorem 3, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 11

Every locally compact metacompact perfectly normal space is paracompact.

Proof of Theorem 11
This follows from Theorem 10 after we prove the following two points:

• Any locally compact space is a Baire space.
• Any perfect locally compact space is locally CCC.

To see the first point, let $Y$ be a locally compact space. Let $W_1,W_2,W_3,\cdots$ be dense open sets in $Y$. Let $y \in Y$ and let $W \subset Y$ be open such that $y \in W$ and $\overline{W}$ is compact. We show that $W$ contains a point that belongs to all $W_n$. Let $X_1=W \cap W_1$, which is open and non-empty. Next choose non-empty open $X_2$ such that $\overline{X_2} \subset X_1$ and $X_2 \subset W_2$. Next choose non-empty open $X_3$ such that $\overline{X_3} \subset X_2$ and $X_3 \subset W_3$. Continue in this manner, we have a sequence of open sets $X_1,X_2,X_3,\cdots$ such that for each $n$, $\overline{X_{n+1}} \subset X_n$ and $\overline{X_n}$ is compact. The intersection of all the $X_n$ is non-empty. The points in the intersection must belong to each $W_n$.

To see the second point, let $Y$ be a locally compact space such that every closed set is a $G_\delta$-set. Suppose that $Y$ is not locally CCC at $y \in Y$. Let $U \subset Y$ be open such that $y \in U$ and $\overline{U}$ is compact. Then $U$ must not have the CCC. Let $\left\{U_\alpha: \alpha<\omega_1 \right\}$ be a pairwise disjoint collection of open subsets of $U$. Let $O=\bigcup \limits_{\alpha<\omega_1} U_\alpha$ and let $C=Y-O$.

Let $C=\bigcap \limits_{n=1}^\infty V_n$ where each $V_n$ is open in $Y$ and $V_{n+1} \subset V_n$ for each integer $n$. For each $\alpha<\omega_1$, pick $y_\alpha \in U_\alpha$. For each $y_\alpha$, there is some integer $f(\alpha)$ such that $y_\alpha \notin V_{f(\alpha)}$. So there must exist some integer $n$ such that $A=\left\{y_\alpha: f(\alpha)=n \right\}$ is uncountable.

The set $A$ is an infinite subset of the compact set $\overline{U}$. So $A$ has a limit point, say $p$ (also called cluster point). Clearly $p \notin O$. So $p \in C$. In particular, $p \in V_n$. Then $V_n$ contains some points of $A$. But for any $y_\alpha \in A$, $y_\alpha \notin V_n=V_{f(\alpha)}$, a contradiction. So $Y$ must be locally CCC at each $y \in Y$. $\blacksquare$

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Some Examples

Example 1
A CCC space $X$ with an uncountable point-finite open covers. This example demonstrates that in Theorem 2, locally-finite or locally-countable cannot be replaced by point-finite. Consider the following product space:

$Y=\prod \limits_{\alpha < \omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$

i.e, the product space of $\omega_1$ many copies of the two-point discrete space $\left\{0,1 \right\}$. Let $X$ be the set of all points $h \in Y$ such that $h(\alpha)=1$ for only finitely many $\alpha<\omega_1$.

The product space $Y$ is the product of separable spaces, hence has the CCC. The space $X$ is dense in $Y$. Hence $X$ has the CCC. For each $\alpha<\omega_1$, define $U_\alpha$ as follows:

$U_\alpha=\left\{h \in X: h(\alpha)=1 \right\}$

Then $\left\{U_\alpha:\alpha<\omega_1 \right\}$ is a point-finite open cover of $X$. Of course, $X$ in this example is not a Baire space. $\blacksquare$

The following three examples center around the four properties in Theorem 7 (Baire + CCC + metacompact imply Lindelof). These examples show that each property in the hypothesis is crucial.

Example 2
A separable non-Lindelof space that is a Baire space. This example shows that the metacompact assumption is crucial for Theorem 7.

The example is the Sorgenfrey plane $S \times S$ where $S$ is the real line with the Sorgenfrey topology (generated by the half-open intervals of the form $[a,b)$). It is well known that $S \times S$ is not Lindelof. The Sorgenfrey plane is Baire and is separable (hence CCC). Furthermore, $S \times S$ is not metacompact (if it were, it would be Lindelof by Theorem 7). $\blacksquare$

Example 3
A non-Lindelof metacompact Baire space $M$. This example shows that the CCC assumption in Theorem 7 is necessary.

This space $M$ is the subspace of Bing’s Example G that has finite support (defined and discussed in the post A subspace of Bing’s example G. It is normal and not collectionwise normal (hence cannot be Lindelof) and metacompact. The space $M$ does not have CCC since it has uncountably many isolated points. Any space with a dense set of isolated points is a Baire space. Thus the space $M$ is also a Baire space. $\blacksquare$

Example 4
A non-Lindelof CCC metacompact non-Baire space $W$. This example shows that the Baire space assumption in Theorem 7 is necessary.

Let $W$ be the set of all non-empty finite subsets of the real line with the Pixley-Roy topology. Note that $W$ is non-Lindelof and has the CCC and is metacompact. Of course it is not Baire. For more information on Pixley-Roy spaces, see the post called Pixley-Roy hyperspaces. For the purpose of this example, the Pixley-Roy space can be built on any uncountable separable metrizable space.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Tall, F. D., The Countable Chain Condition Versus Separability – Applications of Martin’s Axiom, Gen. Top. Appl., 4, 315-339, 1974.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Cartesian Products of Two Paracompact Spaces – Continued

Consider the real line $\mathbb{R}$ with a topology finer than the usual topology obtained by isolating each point in $\mathbb{P}$ where $\mathbb{P}$ is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use $\mathbb{M}$ to denote this topological space. It is a classic result that $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). Even though the Michael line $\mathbb{M}$ is paracompact (it is in fact hereditarily paracompact), $\mathbb{M}$ is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise $\mathbb{M} \times \mathbb{P}$ would be paracompact (hence normal). Result 3 is the statement that if $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal. We also use this result to show that if $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof (see Result 4 below).

This post is a continuation of the post “Cartesian Products of Two Paracompact Spaces”. In that post, four results are listed. They are:

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Result 1 and Result 2 are proved in the previous post “Cartesian Products of Two Paracompact Spaces”. Result 3 and Result 4 are proved in this post. All spaces are assumed to be regular.

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Paracompact Spaces, Lindelof Spaces and Other Information

A paracompact space is one in which every open cover has a locally finite open refinement. The previous post “Cartesian Products of Two Paracompact Spaces” has a basic discussion on paracompact spaces. For the sake of completeness, we repeat here some of the results discussed in that post. A proof of Proposition 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).. For a proof of Proposition 2, see Theorem 3 in the previous post “Cartesian Products of Two Paracompact Spaces”. We provide a proof for Proposition 3.

Proposition 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Proposition 2
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.

Proof of Proposition 3
Let $L$ be a paracompact space. Let $M \subset L$ be a dense Lindelof subspace. Let $\mathcal{U}$ be an open cover of $L$. Since we are working with a regular space, let $\mathcal{V}$ be an open cover of $L$ such that $\left\{\overline{V}: V \in \mathcal{V} \right\}$ refines $\mathcal{U}$. Let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$. Choose $\left\{W_1,W_2,W_3,\cdots \right\} \subset \mathcal{W}$ such that it is a cover of $M$. Since $M \subset \bigcup \limits_{i=1}^\infty W_i$, $\overline{\bigcup \limits_{i=1}^\infty W_i}=L$.

Since the sets $W_i$ come from a locally finite collection, they are closure preserving. Hence we have:

$\overline{\bigcup \limits_{i=1}^\infty W_i}=\bigcup \limits_{i=1}^\infty \overline{W_i}=L$

For each $i$, choose some $U_i \in \mathcal{U}$ such that $\overline{W_i} \subset U_i$. Then $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable subcollection of $\mathcal{U}$ covering the space $L$. $\blacksquare$

A space is said to be a perfectly normal if it is a normal space with the additional property that every closed subset is a $G_\delta$-set in the space (equivalently every open subset is an $F_\sigma$-set). We need two basic results about hereditarily Lindelof spaces. A space is Lindelof if every open cover of that space has a countable subcover. A space is hereditarily Lindelof if every subspace of that space is Lindelof. Proposition 4 below, stated without proof, shows that to prove a space is hereditarily Lindelof, we only need to show that every open subspace is Lindelof.

Proposition 4
Let $L$ be a space. Then $L$ is hereditarily Lindelof if and only if every open subspace of $L$ is Lindelof.

Proposition 5
Let $L$ be a Lindelof space. Then $L$ is hereditarily Lindelof if and only if $L$ is perfectly normal.

Proof of Proposition 5
$\Rightarrow$ Suppose $L$ is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus $L$ is normal. It remains to show that every open subset of $L$ is $F_\sigma$. Let $U \subset L$ be an non-empty open set. For each $x \in U$, let $V_x$ be open such that $x \in V_x$ and $\overline{V_x} \subset U$ (the space is assumed to be regular). By assumption, the open set $U$ is Lindelof. The open sets $V_x$ form an open cover of $U$. Thus $U$ is the union of countably many $\overline{V}_x$.

$\Leftarrow$ Suppose $L$ is perfectly normal. To show that $L$ is hereditarily Lindelof, it suffices to show that every open subset of $L$ is Lindelof (by Proposition 4). Let $U \subset L$ be non-empty open. By assumption, $U=\bigcup \limits_{i=1}^\infty F_i$ where each $F_i$ is a closed set in $L$. Since the Lindelof property is hereditary with respect to closed subsets, $U$ is Lindelof. $\blacksquare$

Another important piece of information that we need is the following metrization theorem. It shows that being a metrizable space is equivalent to have a base that is $\sigma$-locally finite. In proving Result 3, we will assume that the metric factor has such a base. This is a classic metrization theorem (see [1] or [2] or any other standard topology text).

Theorem 6
Let $X$ be a space. Then $X$ is metrizable if and only if $X$ has a $\sigma$-locally finite base.

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Result 3

Result 3 is the statement that:

If $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal.

Result 3 follows from the following two lemmas.

Lemma 7
If the following two conditions hold:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then every open subset of $X \times Y$ is an $F_\sigma$-set in $X \times Y$.

Proof of Lemma 7
Let $U$ be a open subset of $X \times Y$. If $U=\varnothing$, then $U$ is certainly the union of countably many closed sets. So assume $U \ne \varnothing$. Let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$ (by Theorem 6, such a base exists since $Y$ is metrizable).

Consider all non-empty $B \in \mathcal{B}$ such that we can choose nonempty open set $W_B \subset X$ with $W_B \times \overline{B} \subset U$. Since $U$ is non-empty open, such pairs $(B, W_B)$ exist. Let $\mathcal{B}^*$ be the collection of all non-empty $B \in \mathcal{B}$ for which there is a matching non-empty $W_B$. For each $i$, let $\mathcal{B}_i^*=\mathcal{B}^* \cap \mathcal{B}_i$. Of course, each $\mathcal{B}_i^*$ is still locally finite.

Since every open subset of $X$ is an $F_\sigma$-set in $X$, for each $W_B$, we can write $W_B$ as

$W_B=\bigcup \limits_{j=1}^\infty W_{B,j}$

where each $W_{B,i}$ is closed in $X$.

For each $i=1,2,3,\cdots$ and each $j=1,2,3,\cdots$, consider the following collection:

$\mathcal{V}_{i,j}=\left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}$

Each element of $\mathcal{V}_{i,j}$ is a closed set in $X \times Y$. Since $\mathcal{B}_i^*$ is a locally finite collection in $Y$, $\mathcal{V}_{i,j}$ is a locally finite collection in $X \times Y$. Define $V_{i,j}=\bigcup \mathcal{V}_{i,j}$. The set $V_{i,j}$ is a union of closed sets. In general, the union of closed sets needs not be closed. However, $V_{i,j}$ is still a closed set in $X \times Y$ since $\mathcal{V}_{i,j}$ is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:

$\overline{V_{i,j}}=\overline{\bigcup \mathcal{V}_{i,j}}=\overline{\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}}=\bigcup \left\{\overline{W_{B,j} \times \overline{B}}: B \in \mathcal{B}_i^* \right\}$

$=\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}=V_{i,j}$

Finally, we have $U=\bigcup \limits_{i=1}^\infty \bigcup \limits_{j=1}^\infty V_{i,j}$, which is the union of countably many closed sets. $\blacksquare$

Lemma 8
If $X$ is a paracompact space satisfying the following two conditions:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then $X \times Y$ is paracompact.

Proof of Lemma 8
As in the proof of the above lemma, let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Assume that elements of $\mathcal{U}$ are of the form $A \times B$ where $A$ is open in $X$ and $B \in \mathcal{B}$.

For each $B \in \mathcal{B}$, consider the following two items:

$\mathcal{W}_B=\left\{A: A \times B \in \mathcal{U} \right\}$

$W_B=\bigcup \mathcal{W}_B$

To simplify matter, we only consider $B \in \mathcal{B}$ such that $\mathcal{W}_B \ne \varnothing$. Each $W_B$ is open in $X$ and hence by assumption an $F_\sigma$-set in $X$. Thus by Proposition 2, each $W_B$ is paracompact. Note that $\mathcal{W}_B$ is an open cover of $W_B$. Let $\mathcal{H}_B$ be a locally finite open refinement of $\mathcal{W}_B$. Consider the following two items:

For each $j=1,2,3,\cdots$, let $\mathcal{V}_j=\left\{A \times B: A \in \mathcal{H}_B \text{ and } B \in \mathcal{B}_j \right\}$

$\mathcal{V}=\bigcup \limits_{j=1}^\infty \mathcal{V}_j$

We observe that $\mathcal{V}$ is an open cover of $X \times Y$ and that $\mathcal{V}$ refines $\mathcal{U}$. Furthermore each $\mathcal{V}_j$ is a locally finite collection. The open cover $\mathcal{U}$ we start with has a $\sigma$-locally finite open refinement. Thus $X \times Y$ is paracompact. $\blacksquare$

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Result 4

Result 4 is the statement that:

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Proof of Result 4
Suppose $X$ is hereditarily Lindelof and that $Y$ is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus $X$ is paracompact. By Proposition 5, $X$ is perfectly normal. By Result 3, $X \times Y$ is paracompact and perfectly normal.

Let $D$ be a countable dense subset of $Y$. We can think of $D$ as a $\sigma$-compact space. The product of any Lindelof space with a $\sigma$-compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus $X \times D$ is Lindelof. Furthermore $X \times D$ is a dense Lindelof subspace of $X \times Y$. By Proposition 3, $X \times Y$ is Lindelof. By Proposition 5, $X \times Y$ is hereditarily Lindelof. $\blacksquare$

Remark
In the previous post “Bernstein Sets and the Michael Line”, a non-normal product space where one factor is Lindelof and the other factor is a separable metric space is presented. That Lindelof space is not hereditarily Lindelof (it has uncountably many isolated points). Note that by Result 4, for any such non-normal product space, the Lindelof factor cannot be hereditarily Lindelof.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# Cartesian Products of Two Paracompact Spaces

In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking two-factor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

With Results 1 and 2, compact spaces and $\sigma$-compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.

Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.

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Paracompact Spaces

First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a cover of $X$ if $X=\bigcup \mathcal{A}$ (in words every point of the space belongs to one set in the collection). Furthermore, $\mathcal{A}$ is an open cover of $X$ is it is a cover of $X$ consisting of open subsets of $X$.

Let $\mathcal{A}$ and $\mathcal{B}$ be covers of the space $X$. The cover $\mathcal{B}$ is said to be a refinement of $\mathcal{A}$ ($\mathcal{B}$ is said to refine $\mathcal{A}$) if for every $B \in \mathcal{B}$, there is some $A \in \mathcal{A}$ such that $B \subset A$. The cover $\mathcal{B}$ is said to be an open refinement of $\mathcal{A}$ if $\mathcal{B}$ refines $\mathcal{A}$ and $\mathcal{B}$ is an open cover.

A collection $\mathcal{A}$ of subsets of $X$ is said to be a locally finite collection if for each point $x \in X$, there is a non-empty open subset $V$ of $X$ such that $x \in V$ and $V$ has non-empty intersection with at most finitely many sets in $\mathcal{A}$. An open cover $\mathcal{A}$ of $X$ is said to have a locally finite open refinement if there exists an open cover $\mathcal{C}$ of $X$ such that $\mathcal{C}$ refines $\mathcal{A}$ and $\mathcal{C}$ is a locally finite collection. We have the following definition.

Definition

The space $X$ is said to be paracompact if every open cover of $X$ has a locally finite open refinement.

A collection $\mathcal{U}$ of subsets of the space $X$ is said to be a $\sigma$-locally finite collection if $\mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i$ such that each $\mathcal{U}_i$ is a locally finite collection of subsets of $X$. Consider the property that every open cover of $X$ has a $\sigma$-locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following holds:

For each open cover $\left\{U_t: t \in T \right\}$ of $X$, there exists a locally finite open cover $\left\{V_t: t \in T \right\}$ such that $\overline{V_t} \subset U_t$ for each $t \in T$.

Theorem 3 below shows that paracompactness is hereditary with respect to $F_\sigma$-subsets.

Theorem 3
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proof of Theorem 3
Let $X$ be paracompact. Let $Y \subset X$ such that $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is a closed subset of $X$. Let $\mathcal{U}$ be an open cover of $Y$. For each $U \in \mathcal{U}$, let $U^*$ be open in $X$ such that $U^* \cap Y=U$.

For each $i$, let $\mathcal{U}_i^*$ be the set of all $U^*$ such that $U \cap Y_i \ne \varnothing$. Let $\mathcal{V}_i^*$ be a locally finite refinement of $\mathcal{U}_i^* \cup \left\{X-Y_i \right\}$. Let $\mathcal{V}_i$ be the following:

$\mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}$

It is clear that each $\mathcal{V}_i$ is a locally finite collection of open set in $Y$ covering $Y_i$. All the $\mathcal{V}_i$ together form a refinement of $\mathcal{U}$. Thus $\mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. By Theorem 1, the $F_\sigma$-set $Y$ is paracompact. $\blacksquare$
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Result 1

Result 1 is the statement that:

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let $X$ be any space and $Y$ be compact. For each $x \in X$ and for each open set $U \subset X \times Y$ such that $\left\{x \right\} \times Y \subset U$, there is an open set $O \subset X$ such that $\left\{x \right\} \times Y \subset O \times Y \subset U$.

Proof of Result 1
Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, choose a finite $\mathcal{U}_x \subset \mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x \right\} \times Y$. By the Tube Lemma, for each $x \in X$, there is an open set $O_x \subset X$ such that $\left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is paracompact, by Theorem 2, let $\left\{W_x: x \in X \right\}$ be a locally finite open refinement of $\left\{O_x: x \in X \right\}$ such that $W_x \subset O_x$ for each $x \in X$.

Let $\mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}$. We claim that $\mathcal{W}$ is a locally finite open refinement of $\mathcal{U}$. First, this is an open cover of $X \times Y$. To see this, let $(a,b) \in X \times Y$. Then $a \in W_x$ for some $x \in X$. Furthermore, $a \in O_x$ and $(a,b) \in \cup \mathcal{U}_x$. Thus, $(a,b) \in (W_x \times Y) \cap U$ for some $U \in \mathcal{U}_x$. Secondly, it is clear that $\mathcal{W}$ is a refinement of the original cover $\mathcal{U}$.

It remains to show that $\mathcal{W}$ is locally finite. To see this, let $(a,b) \in X \times Y$. Then there is an open $V$ in $X$ such that $x \in V$ and $V$ can meets only finitely many $W_x$. Then $V \times Y$ can meet only finitely many sets in $\mathcal{W}$. $\blacksquare$

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Result 2

Result 2 is the statement that:

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Proof of Result 2
Note that the $\sigma$-compact space $Y$ is Lindelof. Since regular Lindelof are normal, $Y$ is normal and is thus completely regular. So we can embed $Y$ into a compact space $K$. For example, we can let $K=\beta Y$, which is the Stone-Cech compactification of $Y$ (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing $Y$ will do. By Result 1, $X \times K$ is paracompact. Note that $X \times Y$ can be regarded as a subspace of $X \times K$.

Let $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is compact in $Y$. Note that $X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i$ and each $X \times Y_i$ is a closed subset of $X \times K$. Thus the product $X \times Y$ is an $F_\sigma$-subset of $X \times K$. According to Theorem 3, $F_\sigma$-subsets of any paracompact space is paracompact space. Thus $X \times Y$ is paracompact. $\blacksquare$

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# The Michael Line and the Continuum Hypothesis

There exist a Lindelof space and a separable metric space such that their Cartesian product is not normal (discussed in the post “Bernstein Sets and the Michael Line”). The separable metric space is a Bernstein set, a subspace of the real line that is far from being a complete metric space. However, this example is constructed without using any additional set theory axiom beyond the Zermelo-Fraenkel axioms plus the axiom of choice (abbreviated ZFC). A natural question is whether there exists a Lindelof space and a complete metric space such that their product is not normal. In particular, does there exist a Lindelof space $L$ such that the product of $L$ with the space of all irrational numbers is not normal? As of the writing of this post, it is still unknown that such a Lindelof space can exist in just ZFC alone without applying additional set theory axiom. However, such a Lindelof space can be constructed from various additional axioms (e.g. continuum hypothesis or Martin’s axiom). In this post, we present an example of such construction using the continuum hypothesis (the statement that the cardinality of the real line is the same as the first uncountable cardinal $\aleph_1$).

Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace $L$ of $\mathbb{M}$ such that $L$ contains the set $\mathbb{Q}$ of rational numbers. The same proof that $\mathbb{M} \times \mathbb{P}$ is not normal will show that $L \times \mathbb{P}$ is not normal.

See the following posts for a basic discussion of the Michael line:

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Luzin Sets

The Lindelof space $X$ we want to find is a subset of the real line that is called a Luzin set. Before defining Luzin sets, recall some definitions. Let $Y$ be a space. Let $A \subset Y$. The set $A$ is said to be nowhere dense in $Y$ if for every non-empty open subset $U$ of $Y$, there is a non-empty open subset $V$ of $Y$ such that $V \subset U$ and $V$ misses $A$ (equivalently, the closure of $A$ has no interior). The set $A$ is of first category in $Y$ if it is the union of countably many nowhere dense sets.

To define Luzin sets, we focus on the Euclidean space $\mathbb{R}$. Let $A \subset \mathbb{R}$. The set $A$ is said to be a Luzin set if for every set $W \subset \mathbb{R}$ that is of first category in the real line, $A \cap W$ is at most countable. The Russian mathematician Luzin in 1914 constructed such an uncountable Luzin set using continuum hypothesis (CH). A good reference for Luzin sets is [4]. We have the following theorem.

Theorem 1
Assume CH. There exists an uncountable Luzin set.

Proof of Theorem 1
There are continuum many closed nowhere dense subsets of the real line. Since we assume the continuum hypothesis, we can enumerate these sets in a sequence of length $\omega_1$. Let $\left\{F_\alpha: \alpha < \omega_1 \right\}$ be the set of all closed nowhere dense sets in the real line. Choose a real number $x_0 \notin F_0$ to start. For each $\alpha$ with $0 < \alpha <\omega_1$, choose a real number $x_\alpha$ not in the following set:

$\left\{x_\beta: \beta<\alpha \right\} \cup \bigcup \limits_{\beta<\alpha} F_\beta$

The above set is a countable union of closed nowhere dense sets of the real line. As a complete metric space, the real line cannot be of first category. In fact, according to the Baire category theorem, the complement of a set of first category (such as the one described above) is dense in the real line. So such an $x_\alpha$ can always be selected at each $\alpha<\omega_1$. Then $X=\left\{x_\alpha: \alpha<\omega_1 \right\}$ is a Luzin set. $\blacksquare$

Now that we have a way of constructing an uncountable Luzin sets, the following observations provide some useful facts for our problem at hand.

Nowhere dense sets and sets of first category are "thin" sets. Any "thin" set can intersect with a Luzin set with only countably many points. Thus any "co-thin" set contains all but countably many points of a Luzin set. For example, let $A$ be an uncountable Luzin set. Then if $F$ is a closed nowhere dense set in the real line, then $\mathbb{R}-F$ contains all but countably many points of $A$. Furthermore, if $F_1,F_2,F_3,\cdots,$ are closed nowhere dense subsets of the real line, then $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ contains all but countably many points of the Luzin set $A$.

Note that the set $\mathbb{R}-F$ in the preceding paragraph is a dense open set. Thus the complement of a closed nowhere dense set is a dense open set. Note that the set $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ in the preceding paragraph is a dense $G_\delta$-set. Thus the complement of the union of countably many closed nowhere dense sets is a dense $G_\delta$-set. Thus the observation in the preceding paragraph gives the following proposition:

Proposition 2
Given an uncountable Luzin set $A$ and given a dense $G_\delta$ subset $H$ of the real line, $H$ contains all but countably many points of $A$.

In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.

Proposition 3
Let $A$ be an uncountable Luzin set. Let $Y \subset \mathbb{R}$ be uncountable and dense in the real line such that $A \cap Y$ is uncountable. Given a dense $G_\delta$ subset $H$ of $Y$, $H$ contains all but countably many points of $A \cap Y$.

Proof of Proposition 3
We want to show that $Y-H$ can only contain countably many points of $A$. Let $H=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is open and dense in $Y$. Then for each $i$, let $U_i$ be open in the real line such that $U_i \cap Y=O_i$. Each $U_i$ is open and dense in the real line. Thus $H^*=\bigcap \limits_{i=1}^\infty U_i$ contains all but countably many points of the Luzin set $A$. Note the following set inclusion:

$H=\bigcap \limits_{i=1}^\infty U_i \cap Y=\bigcap \limits_{i=1}^\infty O_i \subset \bigcap \limits_{i=1}^\infty U_i=H^*$

Suppose that $Y-H$ contains uncountably many points of $A$. Then these points, except for countably many points, must belong to $H^*=\bigcap \limits_{i=1}^\infty U_i$. The above set inclusion shows that these points must belong to $H$ too, a contradiction. Thus $Y-H$ can only contain countably many points of $A$, equivalently the $G_\delta$-set $H$ contains all but countably many points of $A \cap Y$. $\blacksquare$

The following proposition follows from Proposition 3 and is a useful fact that will help us see that the product of an uncountable Luzin set and $\mathbb{P}$ is not normal.

Proposition 4
Let $Y$ be an uncountable Luzin set such that $\mathbb{Q} \subset Y$. Then $Y-\mathbb{Q}$ cannot be an $F_\sigma$-set in the Euclidean space $Y$, equivalently $\mathbb{Q}$ cannot be a $G_\delta$-set in the space $Y$.

Proof of Proposition 4
By Proposition 3, any dense $G_\delta$-subset of $Y$ must be co-countable. $\blacksquare$

The following proposition is another useful observation about Luzin sets. Let $A \subset \mathbb{R}$. Let $D \subset \mathbb{R}$ be a countable dense subset of the real line. The set $A$ is said to be concentrated about $D$ if for every open subset $O$ of the real line such that $D \subset O$, $O$ contains all but countably many points of $A$. The following proposition can be readily checked based on the definition of Luzin sets.

Proposition 5
For any $A \subset \mathbb{R}$, $A$ is a Luzin set if and only if $A$ is concentrated about every countable dense subset of the real line.

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Lindelof Subspace of The Michael Line

Let $A$ be an uncountable Luzin set. We can assume that $A$ is dense in the real line. If not, just add a countble subset of $\mathbb{P}$ that is dense in the real line. Let $L=A \cup \mathbb{Q}$. It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus $L$ is also a Luzin set. Now consider $L$ as a subspace of the Michael line $\mathbb{M}$. Then points of $L-\mathbb{Q}$ are discrete and points in $\mathbb{Q}$ have Euclidean open neighborhoods. By Proposition 5, the set $L$ is concentrated about every countable dense subset of the real line. In particular, it is concentrated about $\mathbb{Q}$. Thus as a subspace of the Michael line, $L$ is a Lindelof space, since every open set containing $\mathbb{Q}$ contains all but countably many points of $L$.

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The Non-Normal Product $L \times \mathbb{P}$

We highlight the following two facts about the Luzin set $L=A \cup \mathbb{Q}$ as discussed in the preceding section.

• $L-\mathbb{Q}$ is not an $F_\sigma$-set in $L$ (as Euclidean space).
• $A=L-\mathbb{Q}$ is dense in the real line.

The first bullet point follows from Proposition 4. The second bullet point is clear since we assume the Luzin set $A$ we start with is dense. Recall that when thinking of $L$ as a subspace of the Michael line, $L-\mathbb{Q}$ are isolated and $\mathbb{Q}$ retains the usual real line open sets. Because of the above two bullet points, $L \times \mathbb{P}$ is not normal. The proof that $L \times \mathbb{P}$ is not normal is the corollary of the proof that $\mathbb{M} \times \mathbb{P}$ is not normal. Note that in the proof for showing $\mathbb{M} \times \mathbb{P}$ is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an $F_\sigma$-set and are dense in the real line (found in “Michael Line Basics”).

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Michael Space

The example $L \times \mathbb{P}$ that we construct here was hinted in footnote 4 in [6]. In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in [5]). That construction should produce a similar set as the Luzin sets since the approach in [5] is a mirror image of the Luzin set construction. The approach in the Luzin set construction in Theorem 1 is to pick points not in the union of countably many closed nowhere dense sets, while the approach in [5] was to pick points in dense $G_\delta$-sets in a transfinite induction process.

A Michael space is a Lindelof space whose product with $\mathbb{P}$ is not normal. The example shown here shows that under CH, there exists a Michael space. However, the question of whether there exists a Michael space in ZFC is still unsolved. This is called the Michael problem. A recent mention of this unsolved problem is [3] (page 160). A Michael space can also be constructed using Martin’s axiom (see [1]).

A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is $\mathbb{P}$ a productively Lindelof space? As we see here, under CH the answer is no. Another way of looking at the Michael problem: is it possible to show that $\mathbb{P}$ is not productively Lindelof in ZFC alone?

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Reference

1. Alster, K., The product of a Lindelof space with the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., 110 (1990) 543-547.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
5. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
6. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
7. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# Bernstein Sets and the Michael Line

Let $\mathbb{M}$ be the Michael line and let $\mathbb{P}$ be the set of all irrational numbers with the Euclidean topology. In the post called “Michael Line Basics”, we show that the product $\mathbb{M} \times \mathbb{P}$ is not normal. This is a classic counterexample showing that the product of two paracompact spaces need not be normal even when one of the factors is a complete metric space. The Michael line $\mathbb{M}$ is not Lindelof. A natural question is: can the first factor be made a Lindelof space? In this post, as an application of Bernstein sets, we present a non-normal product space where one factor is Lindelof and the other factor is a separable metric space. It is interesting to note that while one factor is upgraded (from paracompact to Lindelof), the other factor is downgraded (from a complete metric space to just a separable metric space).

Bernstein sets have been discussed previously in this blog. They are special subsets of the real line and with the Euclidean subspace topology, they are spaces in which the Banach-Mazur game is undecidable (see the post “Bernstein Sets Are Baire Spaces”). A Bernstein set is a subset $B$ of the real line such that every uncountable closed subset of the real line has non-empty intersection with both $B$ and the complement of $B$.

Bernstein sets are constructed by transfinite induction. The procedure starts by ordering all uncountable closed subsets of the real line in a sequence of length that is as long as the cardinality of continuum. To see how Bernstein sets are constructed, see the post “Bernstein Sets Are Baire Spaces”.

After we discuss a generalization of the definition of the Michael line, we discuss the non-normal product space based on Bernstein sets.
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Generalizing the Michael Line

Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers and let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$. Recall that the Michael line is the real line $\mathbb{R}$ topologized by letting points in $\mathbb{P}$ discrete and letting points in $\mathbb{Q}$ retain their usual open neighborhoods. We can carry out the same process on any partition of the real number line.

Let $D$ and $E$ be disjoint sets such that $\mathbb{R}=D \cup E$ where the set $E$ is dense in the real line. The intention is to make $D$ the discrete part and $E$ the Euclidean part. In other words, we topologize $\mathbb{R}$ be letting points in $D$ discrete and letting points in $E$ retain their Euclidean open sets. Let $X_D$ denote the resulting topological space. For the lack of a better term, we call the space $X_D$ the modified Michael line. An open set in the space $X_D$ is of the form $U \cup V$ where $U$ is a Euclidean open subset of the real line and $V \subset D$. We have the following result:

Proposition
Suppose that $D$ is not an $F_\sigma$-set in the Euclidean real line and that $D$ is dense in the Euclidean real line. Then the product space $X_D \times D$ is not normal (the second factor $D$ is considered a subspace of the Euclidean real line).

In the post “Michael Line Basics”, we give a proof that $\mathbb{M} \times \mathbb{P}$ is not normal. This proof hinges on the same two facts about the set $D$ in the hypothesis in the above proposition. Thus the proof for the above proposition is just like the one for $\mathbb{M} \times \mathbb{P}$. Whenever we topologize the modified Michael line by using a non-$F_\sigma$-set as the discrete part, we can always be certain that we have a non-normal product as indicated here.

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Non-Normal Product Space

Let $B$ be any Bernstein set. The set $B$ is clearly not an $F_\sigma$-set in the real line and is clearly dense in the real line. Then $X_B \times B$ is not normal. Note that in $X_B$, the set $B$ is discrete and its complement $\mathbb{R}-B$ has the usual topology. To see that $X_B$ is Lindelof, note that any open cover of $X_B$ has a countable subcollection that covers $\mathbb{R}-B$. This countable subcollection consists of Euclidean open sets. Furthermore, the complement of the union of these countably many Euclidean open sets must contain all but countably many points of the Bernstein set $B$ (otherwise there would be an uncountable Euclidean closed set that misses $B$).

As commented at the beginning, in obtaining this non-normal product space, one factor is enhanced at the expense of the other factor (one is made Lindelof while the other is no longer a complete metric space). Even though any Bernstein set (with the Euclidean topology) is a separable metric space, it cannot be completely metrizable. Any completely metrizable subset of the real line must be a $G_\delta$-set in the real line. Furthermore any uncountable $G_\delta$ subset of the real line must contain a Cantor set and thus cannot be a Bernstein set.

A similar example to $X_B \times B$ is presented in E. Michael’s paper (see [3]). It is hinted in footnote 4 of that paper that with the additional assumption of continuum hypothesis (CH), one can have a non-normal product space where one factor is a Lindelof space and the second factor is the space of irrationals. So with an additional set-theoretic assumption, we can keep one factor from losing complete metrizability. For this construction, see point (d) in Example 3.2 of [2].

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A Brief Remark

Note that the Lindelof space $X_B$ presented here is not hereditarily Lindelof, since it has uncountably many isolated points. Can a hereditarily Lindelof example be constructed such that its product with a particular separable metric space is not normal? The answer is no. The product of a hereditarily Lindelof space and any separable metric space is hereditarily Lindelof (see Result 4 in the post Cartesian Products of Two Paracompact Spaces – Continued).

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
3. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$