# Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space $X=\omega_1 \times I^I$, which is the product of $\omega_1$ with the interval topology and the product space $I^I=\prod_{t \in I} I$ where $I$ is of course the unit interval $[0,1]$. The notation of $\omega_1$, the first uncountable ordinal, in Steen and Seebach is $[0,\Omega)$.

Another way to notate the example $X$ is the product space $\prod_{t \in I} X_t$ where $X_0$ is $\omega_1$ and $X_t$ is the unit interval $I$ for all $t>0$. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of $\omega_1$ makes this product space an interesting example.

The following lists out the basic topological properties of the space that $X=\omega_1 \times I^I$ are covered in [2].

• The space $X$ is Hausdorff and completely regular.
• The space $X$ is countably compact.
• The space $X$ is neither compact nor sequentially compact.
• The space $X$ is neither separable, Lindelof nor $\sigma$-compact.
• The space $X$ is not first countable.
• The space $X$ is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

• The space $X$ is not normal.
• The space $X$ has a dense subspace that is normal.

It follows from these bullet points that the space $X$ is an example of a completely regular space that is not normal. Not being a normal space, $X$ is then not metrizable. Of course there are other ways to show that $X$ is not metrizable. One is that neither of the two factors $\omega_1$ or $I^I$ is metrizable. Another is that $X$ is not first countable.

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The space $X$ is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example $X=\omega_1 \times I^I$ is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example $X=\omega_1 \times I^I$ is to show whether the second factor $I^I$ is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space $Y$, the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. Thus the normality of the space $X$ (or the lack of) hinges on whether the compact factor $I^I=\prod_{t \in I} I$ is countably tight.

A space $Y$ is countably tight (or has countable tightness) if for each $S \subset Y$ and for each $x \in \overline{S}$, there exists some countable $B \subset S$ such that $x \in \overline{B}$. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space $I^I=\prod_{t \in I} I$ is not countably tight, we let $S$ be the subspace of $I^I$ consisting of points, each of which is non-zero on at most countably many coordinates. Specifically $S$ is defined as follows:

$S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}$

The set $S$ just defined is also called the $\Sigma$-product of copies of unit interval $I$. Let $g \in I^I$ be defined by $g(t)=1$ for all $t \in I$. It follows that $g \in \overline{S}$. It can also be verified that $g \notin \overline{B}$ for any countable $B \subset S$. This shows that the product space $I^I=\prod_{t \in I} I$ is not countably tight.

By Theorem 1 found in this link, the space $X=\omega_1 \times I^I$ is not normal.

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The space $X$ has a dense subspace that is normal

Now that we know $X=\omega_1 \times I^I$ is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace $\omega_1 \times S$ where $S$ is the $\Sigma$-product $S=\Sigma_{t \in I} I$ defined in the preceding section. The subspace $S$ is dense in the product space $I^I$. Thus $\omega_1 \times S$ is dense in $X=\omega_1 \times I^I$. The space $S$ is normal since the $\Sigma$-product of separable metric spaces is normal. Furthermore, $\omega_1$ can be embedded as a closed subspace of $S=\Sigma_{t \in I} I$. Then $\omega_1 \times S$ is homeomorphic to a closed subspace of $S \times S$. Since $S \times S \cong S$, the space $\omega_1 \times S$ is normal.

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Reference

1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Normality in the powers of countably compact spaces

Let $\omega_1$ be the first uncountable ordinal. The topology on $\omega_1$ we are interested in is the ordered topology, the topology induced by the well ordering. The space $\omega_1$ is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of $\omega_1$. We also make comments on normality in the powers of a countably compact non-compact space.

Let $\omega$ be the first infinite ordinal. It is well known that $\omega^{\omega_1}$, the product space of $\omega_1$ many copies of $\omega$, is not normal (a proof can be found in this earlier post). This means that any product space $\prod_{\alpha<\kappa} X_\alpha$, with uncountably many factors, is not normal as long as each factor $X_\alpha$ contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space $\prod_{\alpha<\kappa} X_\alpha$, the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of $\omega$). In other words, the interesting case is that each factor $X_\alpha$ is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in $X^{\kappa}$ where $X$ is a countably non-compact space. In this post we start with the space $X=\omega_1$ of the countable ordinals. We examine $\omega_1$ power $\omega_1^{\omega_1}$ as well as the countable power $\omega_1^{\omega}$. The former is not normal while the latter is normal. The proof that $\omega_1^{\omega}$ is normal is an application of the normality of $\Sigma$-product of the real line.

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The uncountable product

Theorem 1
The product space $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal.

Theorem 1 follows from Theorem 2 below. For any space $X$, a collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for any finite $\mathcal{F} \subset \mathcal{C}$, the intersection $\cap \mathcal{F} \ne \varnothing$. Such a collection $\mathcal{C}$ is called an f.i.p collection for short. It is well known that a space $X$ is compact if and only collection $\mathcal{C}$ of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space $X=\omega_1$, there is an f.i.p. collection of cardinality $\omega_1$ using its linear order. For each $\alpha<\omega_1$, let $C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}$. Let $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$. It is a collection of closed subsets of $X=\omega_1$. It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space $X$ with an f.i.p. collection of cardinality $\omega_1$ that has empty intersection, the product space $X^{\omega_1}$ is not normal.

Theorem 2
Let $X$ be a countably compact space. Suppose that there exists a collection $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ of closed subsets of $X$ such that $\mathcal{C}$ has the finite intersection property and that $\mathcal{C}$ has empty intersection. Then the product space $X^{\omega_1}$ is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space $X^{\omega_1}=\prod_{\alpha<\omega_1} X$, we mean a set of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that each $O_\alpha$ is an open subset of $X$ and that $O_\alpha=X$ for all but finitely many $\alpha<\omega_1$. Given a standard basic open set $O=\prod_{\alpha<\omega_1} O_\alpha$, the notation $\text{Supp}(O)$ refers to the finite set of $\alpha$ for which $O_\alpha \ne X$. For any set $M \subset \omega_1$, the notation $\pi_M$ refers to the projection map from $\prod_{\alpha<\omega_1} X$ to the subproduct $\prod_{\alpha \in M} X$. Each element $d \in X^{\omega_1}$ can be considered a function $d: \omega_1 \rightarrow X$. By $(d)_\alpha$, we mean $(d)_\alpha=d(\alpha)$.

For each $t \in X$, let $f_t: \omega_1 \rightarrow X$ be the constant function whose constant value is $t$. Consider the following subspaces of $X^{\omega_1}$.

$H=\prod_{\alpha<\omega_1} C_\alpha$

$\displaystyle K=\left\{f_t: t \in X \right\}$

Both $H$ and $K$ are closed subsets of the product space $X^{\omega_1}$. Because the collection $\mathcal{C}$ has empty intersection, $H \cap K=\varnothing$. We show that $H$ and $K$ cannot be separated by disjoint open sets. To this end, let $U$ and $V$ be open subsets of $X^{\omega_1}$ such that $H \subset U$ and $K \subset V$.

Let $d_1 \in H$. Choose a standard basic open set $O_1$ such that $d_1 \in O_1 \subset U$. Let $S_1=\text{Supp}(O_1)$. Since $S_1$ is the support of $O_1$, it follows that $\pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_1 \in \bigcap_{\alpha \in S_1} C_\alpha$.

Define $d_2 \in H$ such that $(d_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(d_2)_\alpha=(d_1)_\alpha$ for all $\alpha \in \omega_1-S_1$. Choose a standard basic open set $O_2$ such that $d_2 \in O_2 \subset U$. Let $S_2=\text{Supp}(O_2)$. It is possible to ensure that $S_1 \subset S_2$ by making more factors of $O_2$ different from $X$. We have $\pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_2 \in \bigcap_{\alpha \in S_2} C_\alpha$.

Now choose a point $d_3 \in H$ such that $(d_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(d_3)_\alpha=(d_2)_\alpha$ for all $\alpha \in \omega_1-S_2$. Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

• a sequence $d_1,d_2,d_3,\cdots$ of point of $H=\prod_{\alpha<\omega_1} C_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3 \subset \cdots$ of finite subsets of $\omega_1$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for all $n \ge 2$, $(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $\pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is finite or $A$ is infinite. Let’s examine the two cases.

Case 1
Suppose that $A$ is infinite. Since $X$ is countably compact, $A$ has a limit point $a$. That means that every open set containing $a$ contains some $a_n \ne a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a$ for all $\alpha \in \omega_1-S_n$

From the induction step, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_a \in K$, the constant function whose constant value is $a$. It follows that $t$ is a limit of $\left\{y_1,y_2,y_3,\cdots \right\}$. This means that $t \in \overline{U}$. Since $t \in K \subset V$, $U \cap V \ne \varnothing$.

Case 2
Suppose that $A$ is finite. Then there is some $m$ such that $a_m=a_j$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a_m$ for all $\alpha \in \omega_1-S_n$

As in Case 1, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_{a_m} \in K$, the constant function whose constant value is $a_m$. It follows that $t=y_n$ for all $n \ge m+1$. Thus $U \cap V \ne \varnothing$.

Both cases show that $U \cap V \ne \varnothing$. This completes the proof the product space $X^{\omega_1}$ is not normal. $\blacksquare$

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The countable product

Theorem 3
The product space $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is collectionwise normal, which is stronger than normality. The proof makes use of the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$, which is the following subspace of the product space $\mathbb{R}^{\kappa}$.

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

It is well known that $\Sigma(\kappa)$ is collectionwise normal (see this earlier post). We show that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is a closed subspace of $\Sigma(\kappa)$ where $\kappa=\omega_1$. Thus $\omega_1^{\omega}$ is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space $\omega_1$ is embedded as a closed subspace of $\Sigma(\omega_1)$.

For each $\beta<\omega_1$, define $f_\beta:\omega_1 \rightarrow \mathbb{R}$ such that $f_\beta(\gamma)=1$ for all $\gamma<\beta$ and $f_\beta(\gamma)=0$ for all $\beta \le \gamma <\omega_1$. Let $W=\left\{f_\beta: \beta<\omega_1 \right\}$. We show that $W$ is a closed subset of $\Sigma(\omega_1)$ and $W$ is homeomorphic to $\omega_1$ according to the mapping $f_\beta \rightarrow W$.

First, we show $W$ is closed by showing that $\Sigma(\omega_1)-W$ is open. Let $y \in \Sigma(\omega_1)-W$. We show that there is an open set containing $y$ that contains no points of $W$.

Suppose that for some $\gamma<\omega_1$, $y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}$. Consider the open set $Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1)$ where $Q_\alpha=\mathbb{R}$ except that $Q_\gamma=O$. Then $y \in Q$ and $Q \cap W=\varnothing$.

So we can assume that for all $\gamma<\omega_1$, $y_\gamma \in \left\{0, 1 \right\}$. There must be some $\theta$ such that $y_\theta=1$. Otherwise, $y=f_0 \in W$. Since $y \ne f_\theta$, there must be some $\delta<\gamma$ such that $y_\delta=0$. Now choose the open interval $T_\theta=(0.9,1.1)$ and the open interval $T_\delta=(-0.1,0.1)$. Consider the open set $M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1)$ such that $M_\alpha=\mathbb{R}$ except for $M_\theta=T_\theta$ and $M_\delta=T_\delta$. Then $y \in M$ and $M \cap W=\varnothing$. We have just established that $W$ is closed in $\Sigma(\omega_1)$.

Consider the mapping $f_\beta \rightarrow W$. Based on how it is defined, it is straightforward to show that it is a homeomorphism between $\omega_1$ and $W$.

Claim 2
The $\Sigma$-product $\Sigma(\omega_1)$ has the interesting property it is homeomorphic to its countable power, i.e.

$\Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}$.

Because each element of $\Sigma(\omega_1)$ is nonzero only at countably many coordinates, concatenating countably many elements of $\Sigma(\omega_1)$ produces an element of $\Sigma(\omega_1)$. Thus Claim 2 can be easily verified. With above claims, we can see that

$\displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)$

Thus $\omega_1^{\omega}$ is a closed subspace of $\Sigma(\omega_1)$. Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that $\omega_1^{\omega}$ is normal. $\blacksquare$

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The normality in the powers of $X$

We have established that $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal. Hence any higher uncountable power of $\omega_1$ is not normal. We have also established that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$, the countable power of $\omega_1$ is normal (in fact collectionwise normal). Hence any finite power of $\omega_1$ is normal. However $\omega_1^{\omega}$ is not hereditarily normal. One of the exercises below is to show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let $X$ be a countably compact space has an f.i.p. collection $\mathcal{C}$ of closed sets such that $\bigcap \mathcal{C}=\varnothing$. Then $X^{\kappa}$ is not normal where $\kappa=\lvert \mathcal{C} \lvert$.

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let $X$ be a countably compact space that is not compact. Then there exists a cardinal number $\kappa$ such that $X^{\kappa}$ is not normal and $X^{\tau}$ is normal for all cardinal number $\tau<\kappa$.

By the non-compactness of $X$, there exists an f.i.p. collection $\mathcal{C}$ of closed subsets of $X$ such that $\bigcap \mathcal{C}=\varnothing$. Let $\kappa$ be the least cardinality of such an f.i.p. collection. By Theorem 4, that $X^{\kappa}$ is not normal. Because $\kappa$ is least, any smaller power of $X$ must be normal.

Theorem 6
Let $X$ be a space that is not countably compact. Then $X^{\kappa}$ is not normal for any cardinal number $\kappa \ge \omega_1$.

Since the space $X$ in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, $\omega^{\omega_1}$ is not normal. Then $\omega^{\omega_1}$ is a closed subspace of $X^{\omega_1}$.

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space $X$, $X^{\kappa}$ is not normal for some cardinal number $\kappa$. The $\kappa$ from either Theorem 5 or Theorem 6 is at least $\omega_1$. Interestingly for some spaces, the $\kappa$ can be much smaller. For example, for the Sorgenfrey line, $\kappa=2$. For some spaces (e.g. the Michael line), $\kappa=\omega$.

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space $X$ is normal, then $X$ is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of $X$ is just another way of discussing Theorem 7. According to Theorem 7, if $X$ is not compact, some power of $X$ is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in $X^{\kappa}$ for more examples of countably compact non-compact $X$. One particular interesting example would be a countably compact non-compact $X$ such that the least power $\kappa$ for non-normality in $X^{\kappa}$ is more than $\omega_1$. A possible candidate could be the second uncountable ordinal $\omega_2$. By Theorem 2, $\omega_2^{\omega_2}$ is not normal. The issue is whether the $\omega_1$ power $\omega_2^{\omega_1}$ and countable power $\omega_2^{\omega}$ are normal.

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Exercises

Exercise 1
Show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Exercise 2
Show that the mapping $f_\beta \rightarrow W$ in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space $\omega_1$ is a closed subspace of the $\Sigma$-product of the real line. Show that $\omega_1$ can be embedded in the $\Sigma$-product of arbitrary spaces.

For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Let $p \in \prod_{\alpha<\omega_1} X_\alpha$. The $\Sigma$-product of the spaces $X_\alpha$ is the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$.

$\Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}$

The point $p$ is the center of the $\Sigma$-product. Show that the space $\Sigma(X_\alpha)$ contains $\omega_1$ as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that $\omega_1^{\omega}$ is normal.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Every Corson compact space has a dense first countable subspace

In any topological space $X$, a point $x \in X$ is a $G_\delta$ point if the one-point set $\left\{ x \right\}$ is the intersection of countably many open subsets of $X$. It is well known that any compact Hausdorff space is first countable at every $G_\delta$ point, i.e., if a point of a compact space is a $G_\delta$ point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space $[0,1]^{\omega_1}$ can be a $G_\delta$ point. In this post, we show that any Corson compact space has a dense set of $G_\delta$ point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.

The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.

For any infinite cardinal number $\kappa$, the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$ is the following subspace of $\mathbb{R}^\kappa$:

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

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

For each $x \in \Sigma(\kappa)$, let $S(x)$ denote the support of the point $x$, i.e., $S(x)$ is the set of all $\alpha<\kappa$ such that $x_\alpha \ne 0$.

Proposition 1
Let $Y$ be a Corson compact space. Then $Y$ has a $G_\delta$ point.

Proof of Proposition 1
If $Y$ is finite, then every point is isolated and is thus a $G_\delta$ point. Assume $Y$ is infinite. Let $\kappa$ be an infinite cardinal number such that $Y \subset \Sigma(\kappa)$. For $f,g \in Y$, define $f \le g$ if the following holds:

$\forall \ \alpha \in S(f)$, $f(\alpha)=g(\alpha)$

It is relatively straightforward to verify that the following three properties are satisfied:

• $f \le f$ for all $f \in Y$. (reflexivity)
• For all $f,g \in Y$, if $f \le g$ and $g \le f$, then $f=g$. (antisymmetry)
• For all $f,g,h \in Y$, if $f \le g$ and $g \le h$, then $f \le h$. (transitivity)

Thus $\le$ as defined here is a partial order on the compact space $Y$. Let $C \subset Y$ such that $C$ is a chain with respect to $\le$, i.e., for all $f,g \in C$, $f \le g$ or $g \le f$. We show that $C$ has an upper bound (in $Y$) with respect to the partial order $\le$. We need this for an argument using Zorn’s lemma.

Let $W=\bigcup_{f \in C} S(f)$. For each $\alpha \in W$, choose some $f \in C$ such that $\alpha \in S(f)$ and define $u_\alpha=f_\alpha$. For all $\alpha \notin W$, define $u_\alpha=0$. Because $C$ is a chain, the point $u$ is well-defined. It is also clear that $f \le u$ for all $f \in C$. If $u \in Y$, then $u$ is a desired upper bound of $C$. So assume $u \notin Y$. It follows that $u$ is a limit point of $C$, i.e., every open set containing $u$ contains a point of $C$ different from $u$. Hence $u$ is a limit point of $Y$ too. Since $Y$ is compact, $u \in Y$, a contradiction. Thus it must be that $u \in Y$. Thus every chain in the partially ordered set $(Y,\le)$ has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order $\le$, i.e., there exists $t \in Y$ such that $f \le t$ for all $f \in Y$.

We now show that $t$ is a $G_\delta$ point in $Y$. Let $S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}$. For each $p \in \mathbb{R}$ and for each positive integer $n$, let $B_{p,n}$ be the open interval $B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n})$. For each positive integer $n$, define the open set $O_n$ as follows:

$O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y$

Note that $t \in \bigcap_{n=1}^\infty O_n$. Because $t$ is a maximal element, note that if $g \in Y$ such that $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$, then it must be the case that $g=t$. Thus if $g \in \bigcap_{n=1}^\infty O_n$, then $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$. We have $\left\{t \right\}= \bigcap_{n=1}^\infty O_n$. $\blacksquare$

Lemma 2
Let $Y$ be a compact space such that for every non-empty compact subspace $K$ of $Y$, there exists a $G_\delta$ point in $K$. Then every non-empty open subset of $Y$ contains a $G_\delta$ point.

Proof of Lemma 2
Let $U_1$ be a non-empty open subset of the compact space $Y$. If there exists $y \in U_1$ such that $\left\{y \right\}$ is open in $Y$, then $y$ is a $G_\delta$ point. So assume that every point of $U_1$ is a non-isolated point of $Y$. By regularity, choose an open subset $U_2$ of $Y$ such that $\overline{U_2} \subset U_1$. Continue in the same manner and obtain a decreasing sequence $U_1,U_2,U_3,\cdots$ of open subsets of $Y$ such that $\overline{U_{n+1}} \subset U_n$ for each positive integer $n$. Then $K=\bigcap_{n=1}^\infty \overline{U_n}$ is a non-empty closed subset of $Y$ and thus compact. By assumption, $K$ has a $G_\delta$ point, say $p \in K$.

Then $\left\{p \right\}=\bigcap_{n=1}^\infty W_n$ where each $W_n$ is open in $K$. For each $n$, let $V_n$ be open in $Y$ such that $W_n=V_n \cap K$. For each $n$, let $V_n^*=V_n \cap U_n$, which is open in $Y$. Then $\left\{p \right\}=\bigcap_{n=1}^\infty V_n^*$. This means that $p$ is a $G_\delta$ point in the compact space $Y$. Note that $p \in U_1$, the open set we start with. This completes the proof that every non-empty open subset of $Y$ contains a $G_\delta$ point. $\blacksquare$

Proposition 3
Let $Y$ be a Corson compact space. Then $Y$ has a dense set of $G_\delta$ points.

Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of $Y$ is also Corson compact. By Proposition 1, every compact subspace of $Y$ has a $G_\delta$ point. By Lemma 2, $Y$ has a dense set of $G_\delta$ points. $\blacksquare$

Corollary 4
Every Corson compact space has a dense first countable subspace.

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Blog posts on Corson compact spaces

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Reference

1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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

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

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

# Sigma-products of first countable spaces

A product space is never first countable if there are uncountably many factors. For example, $\prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1}$ is not first countable. In fact any dense subspace of $\mathbb{R}^{\omega_1}$ is not first countable. In particular, the subspace of $\mathbb{R}^{\omega_1}$ consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$ and is denoted by $\Sigma_{\alpha<\omega_1} \mathbb{R}$. However, this $\Sigma$-product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the $\Sigma$-product of first countable spaces is a Frechet space.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $a \in X$. Consider the following subspace of $X$:

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

The above subspace of $X$ is called the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ about the base point $a$. When the base point is understood, we simply say the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ and use the notation $\Sigma_{\alpha \in A} X_\alpha$ to denote the space.

For each $y \in \Sigma_{\alpha \in A} X_\alpha$, define $S(y)$ to be the set of all $\alpha \in A$ such that $y_\alpha \ne a_\alpha$, i.e., the support of the point $y$. Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set $O=\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is an open subset of $X_\alpha$. The support of $O$, denoted by $supp(O)$ is the finite set of all $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

A space $Y$ is said to be first countable if there exists a countable local base at each point in $Y$. A space $Y$ is said to be a Frechet space if for each $y \in Y$ and for each $M \subset Y$, if $y \in \overline{M}$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $M$ such that the sequence converges to $y$. Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

Suppose each factor $X_\alpha$ is a first countable space. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is a Frechet space.

Proof of Theorem 1
Let $\Sigma=\Sigma_{\alpha \in A} X_\alpha$. Let $M \subset \Sigma$ and let $x \in \overline{M}$. We proceed to define a sequence of points $t_n \in M$ such that the sequence $t_n$ converges to $x$. For each $\alpha \in A$, choose a countable local base $\left\{B_{\alpha,j}: j=1,2,3,\cdots \right\}$ at the point $x_\alpha \in X_\alpha$. Assume that $B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots$. Then enumerate the countable set $S(x)$ by $S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}$. Let $C_1=\left\{\beta_{1,1} \right\}$. The following set $O_1$ is an open subset of $\Sigma$.

$O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma$

Note that $O_1$ is an open set containing $x$. Choose $t_2 \in O_1 \cap M$. Enumerate the support $S(t_2)$ by $S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}$. Form the finite set $C_2$ by picking the first two points of $S(x)$ and the first two points of $S(t_2)$, i.e., $C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}$. Then form the following open subset of $\Sigma$.

$O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma$

Choose $t_3 \in O_2 \cap M$. Enumerate the support $S(t_3)$ by $S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}$. Then let $C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}$, i.e., picking the first three points of $S(x)$, the first three points of $S(t_2)$ and the first three points of $S(t_3)$. Now, form the following open subset of $\Sigma$.

$O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma$

Choose $t_4 \in O_2 \cap M$. Let this inductive process continue and we would obtain a sequence $t_2,t_3,t_4,\cdots$ of points of $M$. We claim that the sequence converges to $x$. Before we prove the claim, let’s make a few observations about the inductive process of defining $t_2,t_3,t_4,\cdots$. Let $C=\bigcup_{j=1}^\infty C_j$.

• Each $C_j$ is the support of the open set $O_j$.
• The sequence of open sets $O_j$ is decreasing, i.e., $O_1 \supset O_2 \supset O_3 \supset \cdots$. Thus for each integer $j$, we have $t_k \in O_j$ for all $k \ge j$.
• The support of the point $x$ is contained in $C$, i.e., $S(x) \subset C$.
• The support of the each $t_j$ is contained in $C$, i.e., $S(t_j) \subset C$.
• In fact, $C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots$.
• The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets $O_j$.
• Consequently, for each $j$, $x_\alpha=(t_j)_\alpha=a_\alpha$ for each $\alpha \in A-C$. In other words, $x$ and each $t_j$ agree (and agree with the base point $a$) on the coordinates outside of the countable set $C$.

Let $U=\prod_{\alpha \in A} U_\alpha$ be a standard open set in the product space $X=\prod_{\alpha \in A} X_\alpha$ such that $x \in U$. Let $U^*=U \cap \Sigma$. We show that for some $n$, $t_j \in U^*$ for all $j \ge n$.

Let $F=supp(U)$ be the support of $U$. Let $F_1=F \cap C$ and $F_2=F \cap (A-C)$. Consider the following open set:

$U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma$

Note that $supp(U^{**})=F_1$. For each $\alpha \in F_1$, choose $B_{\alpha,k(\alpha)} \subset U_\alpha$. Let $m$ be the maximum of all $k(\alpha)$ where $\alpha \in F_1$. Then $B_{\alpha,m} \subset U_\alpha$ for each $\alpha \in F_1$. Choose a positive integer $p$ such that:

$F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}$

Let $n=\text{max}(m,p)$. It follows that there exists some $n$ such that $O_n \subset U^{**}$. Then $t_j \in U^{**}$ for all $j \ge n$. It is also the case that $t_j \in U^{*}$ for all $j \ge n$. This is because $x=t_j$ on the coordinates not in $C$. $\blacksquare$

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