# Sigma-products of separable metric spaces are monolithic

Let $\Sigma(\kappa)$ be the $\Sigma$-product of $\kappa$ many copies of the real lines where $\kappa$ is any infinite cardinal number. Any compact space that can be embedded in $\Sigma(\kappa)$ for some $\kappa$ is said to be a Corson compact space. Corson compact spaces play an important role in functional analysis. Corson compact spaces are also very interesting from a topological point of view. Some of the properties of Corson compact spaces are inherited (as subspaces) from the $\Sigma$-product $\Sigma(\kappa)$. One such property is the property that the $\Sigma$-product $\Sigma(\kappa)$ is monolithic, which implies that the closure of any countable subspace of $\Sigma(\kappa)$ is metrizable.

Previous blog posts on $\Sigma$-products:

A previous blog post on monolithic spaces: A short note on monolithic spaces. A listing of other blog posts on Corson compact spaces is given at the end of this post.

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Defining Sigma-product

Let $\kappa$ be an infinite cardinal number. For each $\alpha<\kappa$, let $X_\alpha$ be a topological space. Let $b \in \prod_{\alpha<\kappa} X_\alpha$. The $\Sigma$-product of the spaces $X_\alpha$ about the base point $b$ is defined as follows:

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

If each $X_\alpha=\mathbb{R}$ and if the base point $b$ is such that $b_\alpha=0$ for all $\alpha<\kappa$, then we use the notation $\Sigma(\kappa)$ for $\Sigma_{\alpha<\kappa} X_\alpha$, i.e., $\Sigma(\kappa)$ is defined as follows:

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

A compact space is said to be a Corson compact space if it can be embedded in the $\Sigma$-product $\Sigma(\kappa)$ for some infinite cardinal $\kappa$.

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Monolithic Spaces

A space $X$ is monolithic if for every subspace $Y$ of $X$, the density of $Y$ equals the network weight of $Y$, i.e., $d(Y)=nw(Y)$. A space $X$ is strongly monolithic if for every subspace $Y$ of $X$, the density of $Y$ equals the weight of $Y$, i.e., $d(Y)=w(Y)$. See the previous post called A short note on monolithic spaces.

The proof of the fact that $\Sigma$-product of separable metrizable spaces is monolithic can be worked out quite easily from definitions. Interested readers are invited to walk through the proof. For the sake of completeness, we prove the following theorem.

Theorem 1
Suppose that for each $\alpha<\kappa$, $X_\alpha$ is a separable metric space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} X_\alpha$ is strongly monolithic.

Proof of Theorem 1
Let $b$ be the base point of the $\Sigma$-product $X=\Sigma_{\alpha<\kappa} X_\alpha$. For each $x \in X$, let $S(x)$ be the support of the point $x$, i.e., the set of all $\alpha<\kappa$ such that $x_\alpha \ne b_\alpha$. Let Y be a subspace of $X$. We show that $d(Y)=w(Y)$.

Let $T=\left\{t_\delta: \delta<\tau \right\}$ be a dense subspace of $Y$ such that $d(Y)=\lvert T \lvert=\tau$. Note that $\overline{T}=Y$ (closure is taken in $Y$). Let $S=\bigcup_{\delta<\tau} S(t_\delta)$. Clearly $\lvert S \lvert \le \tau$. Consider the following subspace of $X$:

$X(S)=\left\{x \in X: S(x) \subset S \right\}$

It is clear that $X(S)$ is a closed subspace of $X$. Since $T \subset X(S)$, the closure of $T$ (closure in $X$ or in $Y$) is a subspace of $X(S)$. Thus $Y \subset X(S)$. Note that $\overline{T}=Y \subset X(S)$. Since each $X_\alpha$ has a countable base, the product space $\prod_{\alpha<\tau} X_\alpha$ has a base of cardinality $\tau$. Thus $\prod_{\alpha<\tau} X_\alpha$ has weight $\le \tau$. Since $X(S) \subset \prod_{\alpha<\tau} X_\alpha$, both $Y$ and $X(S)$ have weights $\le \tau$. We have $w(Y) \le d(Y)=\tau$. Note that $d(Y) \le w(Y)$ always holds. Therefore $d(Y)=w(Y)$. $\blacksquare$

Corollary 2
For any infinite cardinal $\kappa$, the $\Sigma$-product $\Sigma(\kappa)$ is strongly monolithic.

Corollary 3
Any Corson compact space is strongly monolithic.

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Blog posts on Corson compact spaces

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

# Revisit a lemma dealing with normality in products of separable metric spaces

In this post we continue to discuss a lemma that has been discussed previously in this post. The lemma characterizes the dense normal subspaces of a product of separable metric spaces. The lemma discussed here has been sharpened over the version in the previous post. Two versions of the lemma are given (Lemma 1 and Lemma 2). Any one of these two versions can be used to prove that the $\Sigma$-product of separable metric spaces is normal (see this blog post).

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. Then the following conditions are equivalent.

1. $Y$ is normal.
2. For any pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$.
3. For any pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$, meaning that $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

The closure in condition 2 and condition 3 is taken in $\pi_B(Y)$. The map $\pi_B$ is the natural projection from the full product space $X=\prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in B} X_\alpha$.

Proof of Lemma 1
$1 \Longrightarrow 2$
Let $H$ and $K$ be disjoint closed subsets of $Y$. Since $Y$ is normal, there exists a continuous function $f: Y \rightarrow [0,1]$ such that $f(H) \subset \left\{0 \right\}$ and $f(H) \subset \left\{1 \right\}$. By Theorem 1 in this previous post, the continuous function $f$ depends on countably many coordinates. This means that there exists a countable $B \subset A$ and there exists a continuous $g:\pi_B(Y) \rightarrow [0,1]$ such that $f= g \circ \pi_B$. The continuity on the full product space is now reduced to the continuity on a countable subproduct. Now $O_H=g^{-1}([0,0.2))$ and $O_K=g^{-1}((0.8,1])$ are disjoint open sets in $\pi_B(Y)$. Since $f= g \circ \pi_B$, it is the case that $\pi_B(H) \subset O_H$ and $\pi_B(K) \subset O_K$. Since $g$ is continuous, we have

$\overline{O_H}=\overline{g^{-1}([0,0.2))} \subset g^{-1}(\overline{[0,0.2)})=g^{-1}([0,0.2]) \ \ \ \ \ \ \ \ (a)$

$\overline{O_K}=\overline{g^{-1}((0.8,1])} \subset g^{-1}(\overline{(0.8,1]})=g^{-1}([0.8,1]) \ \ \ \ \ \ \ \ (b)$

Note that $\overline{\pi_B(H)} \subset \overline{O_H}$ and $\overline{\pi_B(K)} \subset \overline{O_K}$. If $\overline{\pi_B(H)} \cap \overline{\pi_B(K)} \ne \varnothing$, then $g^{-1}([0,0.2]) \cap g^{-1}([0.8,1]) \ne \varnothing$. Thus $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$.

The direction $2 \Longrightarrow 3$ is immediate.

The direction $3 \Longrightarrow 1$ follows from Lemma 1 in this previous post (see the direction $2 \rightarrow 1$ of Lemma 1 in the previous post). $\blacksquare$

The following lemma is another version of Lemma 1 which may be useful in some circumstances. For $B \subset A$, let $\pi_B \times \pi_B$ be the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into $\prod_{\alpha \in B} X_\alpha \times \prod_{\alpha \in B} X_\alpha$ defined by $(\pi_B \times \pi_B)(x,y)=(\pi_B(x),\pi_B(y))$.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. Then the following conditions are equivalent.

1. $Y \times Y$ is normal.
2. For any pair of disjoint closed subsets $H$ and $K$ of $Y \times Y$, there exists a countable $C \subset A$ such that $\overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing$.
3. For any pair of disjoint closed subsets $H$ and $K$ of $Y \times Y$, there exists a countable $C \subset A$ such that $(\pi_C \times \pi_C)(H)$ and $(\pi_C \times \pi_C)(K)$ are separated in $\pi_C(Y) \times \pi_C(Y)$.

Proof of Lemma 2
$1 \Longrightarrow 2$
Let $H$ and $K$ be disjoint closed subsets of $Y \times Y$. Since $Y \times Y$ is normal, there exists a continuous function $f: Y \times Y \rightarrow [0,1]$ such that $f(H) \subset \left\{0 \right\}$ and $f(H) \subset \left\{1 \right\}$. By Theorem 2 in this previous post, the continuous function $f$ depends on countably many coordinates. This means that there exists a countable $C \subset A$ and there exists a continuous $g:\pi_C(Y) \times \pi_C(Y) \rightarrow [0,1]$ such that $f= g \circ (\pi_C \times \pi_C)$. Now $O_H=g^{-1}([0,0.2))$ and $O_K=g^{-1}((0.8,1])$ are disjoint open sets in $\pi_C(Y) \times \pi_C(Y)$. Since $f= g \circ (\pi_C \times \pi_C)$, it is the case that $(\pi_C \times \pi_C)(H) \subset O_H$ and $(\pi_C \times \pi_C)(K) \subset O_K$.

Since $g$ is continuous, conditions (a) and (b) in the proof of Lemma 1 also hold here. Note that $\overline{(\pi_C \times \pi_C)(H)} \subset \overline{O_H}$ and $\overline{(\pi_C \times \pi_C)(K)} \subset \overline{O_K}$. It follows that $\overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing$.

The direction $2 \Longrightarrow 3$ is immediate.

$3 \Longrightarrow 1$
Let $H$ and $K$ be disjoint closed subsets of $Y \times Y$. By condition 3, there exists a countable $C \subset A$ such that $F_H=(\pi_C \times \pi_C)(H)$ and $F_K=(\pi_C \times \pi_C)(K)$ are separated in $M=\pi_C(Y) \times \pi_C(Y)$. Note that $\overline{F_H} \cap F_K=\varnothing$ and $F_H \cap \overline{F_K}=\varnothing$. Consider the following subspace of $M$.

$W=M-\overline{F_H} \cap \overline{F_K}$

The space $W$ is an open subspace of $M$. The space $M$ is a subspace of a product of countably many separable metric spaces. Thus both $M$ and $W$ are also second countable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Both $Cl_W(F_H)$ and $Cl_W(F_K)$ are disjoint closed subsets of $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ with $Cl_W(F_H) \subset G_H$ and $Cl_W(F_K) \subset G_K$. Then $\pi_B^{-1}(G_H) \cap Y$ and $\pi_B^{-1}(G_K) \cap Y$ are disjoint open subsets of $Y$ separating $H$ and $K$. $\blacksquare$

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Remark

The countable sets in both Lemma 1 and Lemma 2 can be expanded to larger countable sets. For example,

for Lemma 1, for any disjoint closed subsets $H$ and $K$ of $Y$:

1. If for some countable set $B$, $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$, then $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$ for any countable $E \subset A$ with $B \subset E$.
2. If for some countable set $B$, $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$, then $\overline{\pi_E(H)} \cap \pi_E(K)=\pi_E(H) \cap \overline{\pi_E(K)}=\varnothing$ for any countable $E \subset A$ with $B \subset E$.

It is straightforward to verify these facts. For the sake of completeness, we verify condition 2. Suppose that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$ for some countable $B \subset A$. Let $E \subset A$ be countable with $B \subset E$. We show $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$. Suppose $x \in \overline{\pi_E(H)} \cap \overline{\pi_E(K)}$. Then $\pi_B(x) \notin \overline{\pi_B(H)} \cap \overline{\pi_B(K)}$. Choose some standard basic open set $O=\prod_{\alpha \in B} O_\alpha$ with $\pi_B(x) \in O$ such that $O \cap \overline{\pi_B(H)}=\varnothing$ and $O \cap \overline{\pi_B(K)}=\varnothing$. Consider $O_1=\prod_{\alpha \in E} O_\alpha$ such that $O_\alpha=X_\alpha$ for all $\alpha \in C-B$. Clearly $x \in O_1$. Then there exist $h \in H$ and $k \in K$ such that $\pi_E(h) \in O_1$ and $\pi_E(h) \in O_1$. It follows that $\pi_B(h) \in O_1$ and $\pi_B(h) \in O$, a contradiction. Thus $\overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing$.

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$\copyright \ 2014 \text{ by Dan Ma}$
Revised 3/31/2014.

# 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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$\copyright \ \ 2012$

# Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.

$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$

The real line with the topology generated by $\mathcal{B}$ is called the Michael line and is denoted by $\mathbb{M}$. In essense, in $\mathbb{M}$, points in $\mathbb{P}$ are made isolated and points in $\mathbb{Q}$ retain the usual Euclidean open sets.

The Euclidean topology $\tau$ is coarser (weaker) than the Michael line topology (i.e. $\tau$ being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, $\mathbb{M}$ is submetrizable (submetrized by the Euclidean topology). It is clear that $\mathbb{M}$ is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

1. The space $\mathbb{M}$ is paracompact.
2. The space $\mathbb{M}$ is not Lindelof.
3. The extent of the space $\mathbb{M}$ is $c$ where $c$ is the cardinality of the real line.
4. The space $\mathbb{M}$ is not locally compact.
5. The space $\mathbb{M}$ is not perfectly normal, thus not metrizable.
6. The space $\mathbb{M}$ is not a Moore space, but has a $G_\delta$-diagonal.
7. The product $\mathbb{M} \times \mathbb{P}$ is not normal where $\mathbb{P}$ has the usual topology.
8. The product $\mathbb{M} \times \mathbb{P}$ is metacompact.
9. The space $\mathbb{M}$ has a point-countable base.
10. For each $n=1,2,3,\cdots$, the product $\mathbb{M}^n$ is paracompact.
11. The product $\mathbb{M}^\omega$ is not normal.
12. There exist a Lindelof space $L$ and a separable metric space $W$ such that $L \times W$ is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on $\mathbb{Q}$ coincides with the Euclidean topology on $\mathbb{Q}$. A set is said to be a $G_\delta$-set if it is the intersection of countably many open sets. By the Baire category theorem, the set $\mathbb{Q}$ is not a $G_\delta$-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set $\mathbb{Q}$ is not a $G_\delta$-set in the Michael line. This fact is used in Result 5.

The fact that $\mathbb{Q}$ is not a $G_\delta$-set in the Euclidean real line implies that $\mathbb{P}$ is not an $F_\sigma$-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let $\mathcal{U}$ be an open cover of $\mathbb{M}$. We proceed to derive a locally finite open refinement $\mathcal{V}$ of $\mathcal{U}$. Recall that $\tau$ is the usual topology on $\mathbb{R}$. Assume that $\mathcal{U}$ consists of open sets in the base $\mathcal{B}$. Let $\mathcal{U}_\tau=\mathcal{U} \cap \tau$. Let $Y=\cup \mathcal{U}_\tau$. Note that $Y$ is a Euclidean open subspace of the real line (hence it is paracompact). Then there is $\mathcal{V}_\tau \subset \tau$ such that $\mathcal{V}_\tau$ is a locally finite open refinement $\mathcal{V}_\tau$ of $\mathcal{U}_\tau$ and such that $\mathcal{V}_\tau$ covers $Y$ (locally finite in the Euclidean sense). Then add to $\mathcal{V}_\tau$ all singleton sets $\left\{ x \right\}$ where $x \in \mathbb{M}-Y$ and let $\mathcal{V}$ denote the resulting open collection.

The resulting $\mathcal{V}$ is a locally finite open collection in the Michael line $\mathbb{M}$. Furthermore, $\mathcal{V}$ is also a refinement of the original open cover $\mathcal{U}$. $\blacksquare$

A similar argument shows that $\mathbb{M}$ is hereditarily paracompact.

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Result 2

To see that $\mathbb{M}$ is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in $\mathbb{P}$). For example, it is possible to construct a Cantor set entirely within $\mathbb{P}$.

Let $C$ be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set $C$ is an uncountable closed and discrete set in $\mathbb{M}$. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line $\mathbb{M}$ cannot be Lindelof. $\blacksquare$

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space $X$ is the smallest infinite cardinal number $\mathcal{K}$ such that every closed and discrete set in $X$ has cardinality $\le \mathcal{K}$. The extent of the space $X$ is denoted by $e(X)$. When the cardinal number $e(X)$ is $e(X)=\aleph_0$ (the first infinite cardinal number), the space $X$ is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When $e(X)>\aleph_0$, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space $X$ is Lindelof, the extent is $e(X)=\aleph_0$. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of $\mathbb{M}$ of cardinality $c$. Thus we have $e(\mathbb{M})=c$. $\blacksquare$

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Result 4

The Michael line $\mathbb{M}$ is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in $\mathbb{M}$. $\blacksquare$

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Result 5

A set is said to be a $G_\delta$-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a $G_\delta$-set. In the Michael line $\mathbb{M}$, the set $\mathbb{Q}$ of rational numbers is a closed set. Yet, $\mathbb{Q}$ is not a $G_\delta$-set in the Michael line (see the discussion above on the Baire category theorem). Thus $\mathbb{M}$ is not perfectly normal and hence not a metrizable space. $\blacksquare$

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Result 6

The diagonal of a space $X$ is the subset of its square $X \times X$ that is defined by $\Delta=\left\{(x,x): x \in X \right\}$. If the space is Hausdorff, the diagonal is always a closed set in the square. If $\Delta$ is a $G_\delta$-set in $X \times X$, the space $X$ is said to have a $G_\delta$-diagonal. It is well known that any metric space has $G_\delta$-diagonal. Since $\mathbb{M}$ is submetrizable (submetrized by the usual topology of the real line), it has a $G_\delta$-diagonal too.

Any Moore space has a $G_\delta$-diagonal. However, the Michael line is an example of a space with $G_\delta$-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus $\mathbb{M}$ is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. $\blacksquare$

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Result 7

We now show that $\mathbb{M} \times \mathbb{P}$ is not normal where $\mathbb{P}$ has the usual topology. In this proof, the following two facts are crucial:

• The set $\mathbb{P}$ is not an $F_\sigma$-set in the real line.
• The set $\mathbb{P}$ is dense in the real line.

Let $H$ and $K$ be defined by the following:

$H=\left\{(x,x): x \in \mathbb{P} \right\}$
$K=\mathbb{Q} \times \mathbb{P}$.

The sets $H$ and $K$ are disjoint closed sets in $\mathbb{M} \times \mathbb{P}$. We show that they cannot be separated by disjoint open sets. To this end, let $H \subset U$ and $K \subset V$ where $U$ and $V$ are open sets in $\mathbb{M} \times \mathbb{P}$.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval $(a,b)$, we mean the set of all real numbers $t$ with $a. By $(a,b)^*$, we mean $(a,b) \cap \mathbb{P}$. For each $x \in \mathbb{P}$, choose an open interval $U_x=(a,b)^*$ such that $\left\{x \right\} \times U_x \subset U$. We also assume that $x$ is the midpoint of the open interval $U_x$. For each positive integer $k$, let $P_k$ be defined by:

$P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}$

Note that $\mathbb{P}=\bigcup \limits_{k=1}^\infty P_k$. For each $k$, let $T_k=\overline{P_k}$ (Euclidean closure in the real line). It is clear that $\bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k$. On the other hand, $\bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P}$ (otherwise $\mathbb{P}$ would be an $F_\sigma$-set in the real line). So there exists $T_n=\overline{P_n}$ such that $\overline{P_n} \not\subset \mathbb{P}$. So choose a rational number $r$ such that $r \in \overline{P_n}$.

Choose a positive integer $j$ such that $\frac{2}{j}<\frac{1}{n}$. Since $\mathbb{P}$ is dense in the real line, choose $y \in \mathbb{P}$ such that $r-\frac{1}{j}. Now we have $(r,y) \in K \subset V$. Choose another integer $m$ such that $\frac{1}{m}<\frac{1}{j}$ and $(r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V$.

Since $r \in \overline{P_n}$, choose $x \in \mathbb{P}$ such that $r-\frac{1}{m}. Now it is clear that $(x,y) \in V$. The following inequalities show that $(x,y) \in U$.

$\lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}$

The open interval $U_x$ is chosen to have length $> \frac{1}{n}$. Since $\lvert x-y \lvert < \frac{1}{n}$, $y \in U_x$. Thus $(x,y) \in \left\{ x \right\} \times U_x \subset U$. We have shown that $U \cap V \ne \varnothing$. Thus $\mathbb{M} \times \mathbb{P}$ is not normal. $\blacksquare$

Remark
As indicated above, the proof of Result 7 hinges on two facts about $\mathbb{P}$, namely that it is not an $F_\sigma$-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set $D$ that is isolated is not an $F_\sigma$-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space $D$ (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product $\mathbb{M} \times \mathbb{P}$ is not paracompact since it is not normal. However, $\mathbb{M} \times \mathbb{P}$ is metacompact.

A collection of subsets of a space $X$ is said to be point-finite if every point of $X$ belongs to only finitely many sets in the collection. A space $X$ is said to be metacompact if each open cover of $X$ has an open refinement that is a point-finite collection.

Note that $\mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P})$. The first $\mathbb{P}$ in $\mathbb{P} \times \mathbb{P}$ is discrete (a subspace of the Michael line) and the second $\mathbb{P}$ has the Euclidean topology.

Let $\mathcal{U}$ be an open cover of $\mathbb{M} \times \mathbb{P}$. For each $a=(x,y) \in \mathbb{Q} \times \mathbb{P}$, choose $U_a \in \mathcal{U}$ such that $a \in U_a$. We can assume that $U_a=A \times B$ where $A$ is a usual open interval in $\mathbb{R}$ and $B$ is a usual open interval in $\mathbb{P}$. Let $\mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace$.

Fix $x \in \mathbb{P}$. For each $b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}$, choose some $U_b \in \mathcal{U}$ such that $b \in U_b$. We can assume that $U_b=\lbrace{x}\rbrace \times B$ where $B$ is a usual open interval in $\mathbb{P}$. Let $\mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace$.

As a subspace of the Euclidean plane, $\bigcup \mathcal{G}$ is metacompact. So there is a point-finite open refinement $\mathcal{W}$ of $\mathcal{G}$. For each $x \in \mathbb{P}$, $\mathcal{H}_x$ has a point-finite open refinement $\mathcal{I}_x$. Let $\mathcal{V}$ be the union of $\mathcal{W}$ and all the $\mathcal{I}_x$ where $x \in \mathbb{P}$. Then $\mathcal{V}$ is a point-finite open refinement of $\mathcal{U}$.

Note that the point-finite open refinement $\mathcal{V}$ may not be locally finite. The vertical open intervals in $\lbrace{x}\rbrace \times \mathbb{P}$, $x \in \mathbb{P}$ can “converge” to a point in $\mathbb{Q} \times \mathbb{P}$. Thus, metacompactness is the best we can hope for. $\blacksquare$

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base $\mathcal{G}$ for a space $X$ is said to be a point-countable base if $\mathcal{G}$, in addition to being a base for the space $X$, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for $\mathbb{M}$:

$\mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$

where $\mathcal{H}$ is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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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 Evaluation Map

The evaluation map is a useful tool for embedding a space $X$ into a product space. In this post we demonstrate that any Tychonoff space $X$ can be embedded into a cube $I^{\mathcal{K}}$ where $I$ is the unit interval $[0,1]$ and $\mathcal{K}$ is some cardinal. Any regular space with a countable base (second-countable space) can also be embedded into the Hilbert cube $I^{\omega}$ (Urysohn’s metrization theorem). The evaluation map also plays an important role in the theory of Cech-Stone compactification.

The Evaluation Map
Let $X$ be a space. Let $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$ be a product space. For each $y \in Y$, we use the notation $y=\langle y_\alpha \rangle_{\alpha \in A}$ to denote a point in the product space $Y$. Suppose we have a family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ where $f_\alpha:X \rightarrow Y_\alpha$ for each $\alpha$. Define a mapping $E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha$ as follows:

For each $x \in X$, $E_{\mathcal{F}}(x)$ is the point $\langle f_\alpha(x) \rangle_{\alpha \in A} \in Y$.

This mapping is called the evaluation map of the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$. If the family $\mathcal{F}$ is understood, we may skip the subscript and use $E$ to denote the evaluation map.

The family of continuous functions $\mathcal{F}$ is said to separate points if for any two distinct points $x,y \in X$, there is a function $f \in \mathcal{F}$ such that $f(x) \neq f(y)$. The family of continuous functions $\mathcal{F}$ is said to separate points from closed sets if for each point $x \in X$ and for each closed set $C \subset X$ with $x \notin C$, there is a function $f \in \mathcal{F}$ such that $f(x) \notin \overline{f(C)}$.

Theorem 1. Given an evaluation map $E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha$ as defined above, the following conditions hold.

1. The mapping $E_{\mathcal{F}}$ is continuous.
2. If the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ separates points, then $E_{\mathcal{F}}$ is a one-to-one map.
3. If the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ separates points from closed sets, then $E_{\mathcal{F}}$ is a homeomorphism from $X$ into the product space $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$.

In this post, basic open sets in the product space $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$ are of the form $\bigcap_{\alpha \in W} [\alpha,V_\alpha]$ where $W \subset A$ is finite, for each $\alpha \in W$, $V_\alpha$ is an open set in $Y_\alpha$ and $[\alpha,V_\alpha]=\lbrace{y \in Y:y_\alpha \in V_\alpha}\rbrace$.

Proof of 1. We show that $E_{\mathcal{F}}$ is continuous at each $x \in X$. Let $x \in X$. Let $h=\langle f_\alpha(x) \rangle_{\alpha \in A}$ and let $h \in V \cap E_{\mathcal{F}}(X)$ where $V=\bigcap_{\alpha \in W} [\alpha,V_\alpha]$ is a basic open set. Consider $U=\bigcap_{\alpha \in W} f_\alpha^{-1}(V_\alpha)$. It is easy to verify that $x \in U$ and $E_{\mathcal{F}}(U) \subset V\cap E_{\mathcal{F}}(X)$.

Proof of 2. Let $x,y \in X$ be distinct points. There is $\alpha \in A$  such that $f_\alpha(x) \neq f_\alpha(y)$. Clearly, $E_{\mathcal{F}}(x)= \langle f_\beta(x) \rangle_{\beta \in A} \neq E_{\mathcal{F}}(y)=\langle f_\beta(y) \rangle_{\beta \in A}$.

Proof of 3. Note that by condition 2 in this theorem, the map $E_{\mathcal{F}}$ is one-to-one. It suffices to show that $E_{\mathcal{F}}$ is an open map. Let $U \subset X$ be open. We show that $E_{\mathcal{F}}(U)$ is open in $E_{\mathcal{F}}(X)$. To this end, let $\langle f_\alpha(x) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U)$. Then $x \in U$. Since $\mathcal{F}$ separates points from closed sets, there is some $\beta$ such that $f_\beta(x) \notin \overline{f_\beta(X-U)}$. Let $V_\beta=Y_\beta-\overline{f_\beta(X-U)}$. Then $\langle f_\alpha(x) \rangle_{\alpha \in A} \in [\beta,V_\beta] \cap E_{\mathcal{F}}(X)=W_\beta$. We show that $W_\beta \subset E_{\mathcal{F}}(U)$. For each $\langle f_\alpha(y) \rangle_{\alpha \in A} \in W_\beta$, we have $f_\beta(y) \notin \overline{f_\beta(X-U)}$. If $y \notin U$, then $f_\beta(y) \in f_\beta(X-U)$, a contradiction. So we have $y \in U$ and this means that $\langle f_\alpha(y) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U)$. It follows that $W_\beta \subset E_{\mathcal{F}}(U)$.

Some Applications

A space $X$ is a Tychonoff space (also known as completely regular space) if for each $x \in X$ and for each closed set $C \subset X$ where $x \notin C$, there is a continuous function $f:X \rightarrow I$ such that $f(x)=1$ and $f(y)=0$ for all $y \in C$. The following is a corollary to theorem 1.

Corollary 1. Any Tychonoff space can be embedded in a cube $I^{\mathcal{K}}$.

Proof. Let $\mathcal{F}$ be the family of all continuous functions from the Tychonoff space $X$ into the unit interval $I$. By the definition of Tychonoff space, $\mathcal{F}$ separates points from closed sets. By theorem 1, the evaluation map $E_{\mathcal{F}}$ is a homeomorphism from $X$ into the cube $I^{\mathcal{K}}$ where $\mathcal{K}=\lvert \mathcal{F} \lvert$.

We now turn our attention to regular second countable space. Having a countable base has many strong properties, one of which is that it can be embedded into the Hilbert Cube $I^{\omega}=I^{\aleph_0}$. Before we prove this, observe that any regular space with a countable base is a regular Lindelof space. Furthermore, the property of having a countable base is hereditary. Thus a regular space with a countable base is hereditarily Lindelof (hence perfectly normal). The Vendenisoff Theorem states that in a perfectly normal space, every closed set is a zero-set (i.e. every open set is a cozero-set). So we make use of this theorem to obtain continuous functions that separate points from closed sets. There is a proof of The Vendenisoff Theorem in this blog. A set $Z \subset X$ is a zero-set in the space $X$ if there is a continuous function $f:X \rightarrow I$ such that $f^{-1}(0)=Z$. A set $W \subset X$ is a cozero-set if $X-W$ is a zero-set. We are now ready to prove one part of the Urysohn’s metrization theorem.

Urysohn’s metrization theorem. The following conditions are equivalent.

1. The space $X$ is a regular space with a countable base.
2. The space $X$ can be embedded into the Hilbert cube $I^{\aleph_0}$.
3. The space $X$ is a separable metric space.

We prove the direction $1 \Rightarrow 2$. Let $\lbrace{B_0,B_1,B_2,...}\rbrace$ be a countable base for the regular space $X$. Based on the preceding discussion, $X$ is perfectly normal. By the Vendenisoff Theorem, for each $n$$X-B_n$ is a zero-set. Thus for each $n$, there is a continuous function $f_n:X \rightarrow I$ such that $f_n^{-1}(0)=X-B_n$ and $f_n^{-1}((0,1])=B_n$. Let $\mathcal{F}=\lbrace{f_0,f_1,f_2,...}\rbrace$. It is easy to verify that $\mathcal{F}$ separates points from closed sets. Thus the evaluation map $E_{\mathcal{F}}$ is a homeomorphism from $X$ into $I^{\aleph_0}$.

# Perfect Image of Separable Metric Spaces

In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping $f:X \rightarrow Y$ is a closed mapping if $f(H)$ is closed in $Y$ for any closed set $H \subset X$. A continuous surjection $f:X \rightarrow Y$ is a perfect mapping if $f$ is closed and $f^{-1}(y)$ is compact for each $y \in Y$.

Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. Because $f$ is a closed mapping, $f(X-B)$ is closed and $f(B)$ is open in $Y$ for each $B \in \mathcal{B}$. We show that $\mathcal{B}_f=\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$. Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. We have $y \in f(B) \subset U$. Thus the topology on $Y$ can be generated by $\mathcal{B}_f$.

Update (11/24/2009):
The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

Lemma. Let $f: X \rightarrow Y$ be a closed mapping and let $V \subset X$ be open. Then $f_*(V)=\lbrace{y \in Y:f^{-1}(y) \subset V}\rbrace$ is open in $Y$. Furthermore, $f_*(V) \subset f(V)$.

Proof of Lemma. Since $f$ is a closed mapping, $f(X-V)$ is closed. We claim that $f(X-V)=Y-f_*(V)$. It is clear that $f(X-V) \subset Y-f_*(V)$. To show that $Y-f_*(V) \subset f(X-V)$, let $z \in Y-f_*(V)$. Then $f^{-1}(z)$ cannot be a subset of $V$. Choose $x \in f^{-1}(z)-V$. Then we have $z=f(x) \in f(X-V)$. Thus $f(X-V)=Y-f_*(V)$ and $f_*(V)$ is open. It is straitforward to verify that $f_*(V) \subset f(V)$.

Now I prove that the perfect image of a separable metric space is a separable metric space. Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_f=\lbrace{f_*(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$.

Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. Since $f^{-1}(y) \subset B$, we have $y \in f_*(B)$. We also have $f_*(B) \subset f(B) \subset U$. Thus the topology on $Y$ can be generated by the countable base $\mathcal{B}_f$.