# Bing’s G

This post is a basic discussion of Bing’s Example G. The original post was published on 10/27/2009 and is now replaced by a new post. The following link will take you there. Thank you.

# When Is A Product of Metrizable Spaces Normal?

Let $\mathbb{N}$ be the discrete space the natural numbers. It is known that $\mathbb{N}^{\omega_1}$ is not normal (see a proof here). It turns out that the product of uncountably many non-compact metric spaces is never normal. In 1948, A. H. Stone proved that the uncountable product of metric spaces is normal if and only if all but countably many factors are compact. Thus product of uncountably many copies of $\mathbb{R}$ is never normal.

Theorem. Let $\lbrace{Y_\alpha:\alpha \in S}\rbrace$ be a family of metrizable spaces. The following conditions are equivalent.
(1) $\Pi_{\alpha \in S}Y_\alpha$ is paracompact.
(2) $\Pi_{\alpha \in S}Y_\alpha$ is normal.
(3) All but countably many $Y_\alpha$ are compact.

Proof. $(1) \rightarrow (2)$ is obvious.

$(2) \rightarrow (3)$. Suppose $Y=\Pi_{\alpha \in S}Y_\alpha$ is normal. I Claim that all but countably many factors are countably compact. When this claim is established, (3) is established. Note that if a paracompact space is countably compact, it is compact. Suppose that there are uncountably many $Y_\alpha$ that are not countably compact where $\alpha < \omega_1$. Then each such $Y_\alpha$ would contain a closed copy of $\mathbb{N}$. Thus the product space $Y$ would contains $\mathbb{N}^{\omega_1}$ as a closed subspace. This is a contradiction since any closed subspace of a normal space is normal. Thus all but countably many factors are countably compact.

$(3) \rightarrow (1)$. Suppose all but countably many $Y_\alpha$ are compact. Then $Y=\Pi_{\alpha \in S}Y_\alpha=H \times G$ where $H$ is the product of all the compact factors $Y_\alpha$ and $G$ is the product of the countably many non-compact factors. Note that $G$ is also a metrizable space. The product of a compact space and a paracompact space is paracompact (see a proof here). Thus (1) is established.

Corollary. Let $\lbrace{Y_\alpha:\alpha \in S}\rbrace$ be a family of separable metrizable spaces. The following conditions are equivalent.
(1) $\Pi_{\alpha \in S}Y_\alpha$ is paracompact.
(2) $\Pi_{\alpha \in S}Y_\alpha$ is Lindelof.
(3) $\Pi_{\alpha \in S}Y_\alpha$ is normal.
(4) All but countably many $Y_\alpha$ are compact.

The only thing I want to mention about the corollary is that being a product of separable spaces, the product space $\Pi_{\alpha \in S}Y_\alpha$ has the countable chain condition (ccc). In any space with the ccc, paracompactness implies the Lindelof property (see a proof here).

Reference
[Stone] Stone, A. H., [1948] Paracompact and Product Spaces, Bull. Amer. Math. Soc., 54, 977-982.

# Compact x Paracompact Is Paracompact

The theorem indicated by the title is a basic result and is an interesting theorem. It is known that “compact x normal” needs not be normal (see an example here). But for a compact factor and a paracompact factor, the product is not only normal, it is paracompact. I would like to write down a proof. This theorem will be used in subsequent posts. If the factor $Y$ is Lindelof instead, it can be shown that $X \times Y$ is Lindelof.

Let $X$ be a compact space and let $Y$ be a paracompact space. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $y \in Y$, let $\mathcal{G}_y \subset \mathcal{U}$ be a finite cover of $X \times \lbrace{y}\rbrace$.

Claim. Fix $y \in Y$. There is an open set $W_y \subset Y$ such that $X \times W_y \subset \cup \mathcal{G}_y$. For each $x \in X$, let $A_x \subset X$ be open and let $B_x \subset Y$ be open such that $x \in A_x$, $y \in B_x$, and $A_x \times B_x \subset \cup\mathcal{G}_y$. Since $X$ is compact, choose finitely many open sets $A_x$ that cover $X$, say $A_{x(0)},A_{x(1)},...,A_{x(n)}$. Let $W_y=B_{x(0)} \cap B_{x(1)} \cap...\cap B_{x(n)}$. It is clear that $X \times W_y \subset \cup \mathcal{G}_y$.

The collection $\mathcal{W}$$=\lbrace{W_y:y \in Y}\rbrace$ is an open cover of $Y$. Then it has a locally finite open refinement $\mathcal{E}$. For each $E \in$$\mathcal{E}$, choose $y \in Y$ such that $E \subset W_y$. Consider $(X \times E) \cap G$ where $G \in$$\mathcal{G}_y$. Let $\mathcal{F}$ be the collection of all such open sets $(X \times E) \cap G$.

There are three things to check here. One is that $\mathcal{F}$ is an open cover of $X \times Y$. Fix $(x,z) \in X \times Y$. Then $z \in E$ for some $E \in$$\mathcal{E}$. Note that $X \times E \subset X \times W_y \subset \cup \mathcal{G}_y$ for some $y \in Y$. This means that the point $(x,z) \in (X \times E)\cap G$ for some $G$ in $\mathcal{G}_y$.

The second to check is that $\mathcal{F}$ is a refinement of the original open cover $\mathcal{U}$. This is clear since every set in $\mathcal{F}$ is chosen to be a subset of some set in $\mathcal{G}_y \subset \mathcal{U}$.

The third is that $\mathcal{F}$ is a locally finite collection. To see this, let $(x,z) \in X \times Y$. There is an open $V \subset Y$ such that $z \in V$ and $V$ can only meet finitely many $E \in$$\mathcal{E}$. For each such $E$, $X \times E$ is associated with finitely many sets in $\mathcal{F}$. This means $X \times V$ is an open set containing $(x,z)$ that can meet only finitely many sets in $\mathcal{F}$. This completes the proof that $X \times Y$ is paracompact.

If the factor Y is Lindelof, then we can modify the proof to show that $X \times Y$ is Lindelof. In the step above where $\mathcal{W}$$=\lbrace{W_y:y \in Y}\rbrace$ is obtained, we get a countable subcover of $\mathcal{W}$. Each member of this countable subcover is associated with a finite $\mathcal{G}_y \subset \mathcal{U}$. Thus we can obtain a countable subcover of $\mathcal{U}$.

# No Hereditarily Normalilty In Uncountable Product

In a previous post, I showed that the uncountable product of the space of the integers is not normal. Consequently, the uncountable product of the unit interval is never hereditarily normal since it contains a copy of the uncountable product of the integers. It turns out that, as the title of this post suggests, hereditarily normality cannot happen in uncountable product of spaces. Specifically the product of uncountably many spaces, each of which has at least two points, is never hereditarily normal.

Let $Y=\Pi_{\alpha \in A} X_\alpha$ such that $\vert A \vert \geq \omega_1$ and each $X_\alpha$ has at least two points. The proof is quite simple, a matter of rearranging the factors so that the product space $Y$ contains a copy of $\Pi_{\alpha \in A} \mathbb{N}$.

For each $\alpha \in A$, let $Z_\alpha \subset X_\alpha$ be a two-point subspace. Break up $A$ into $\vert A \vert$ many disjoint subsets $A_z$, $z \in A$ where $\vert A_z \vert=\omega$. For each $z \in A$, let $C_z=\Pi_{h \in A_z} Z_h$, which is homeomorphic to the middle third Cantor set in the unit interval. Thus, $\Pi_{\alpha \in A}Z_\alpha=\Pi_{z \in A}C_z \subset \Pi_{\alpha \in A}X_\alpha$.

Each Cantor set contains a subset that is homeomorphic to the discrete space of the natural numbers $\mathbb{N}$. Thus $\Pi_{z \in A}\mathbb{N} \subset \Pi_{z \in A}C_z$.

# The Tychonoff Plank

The Tychonoff plank is the product space $[0,\omega_1] \times [0,\omega]$, that is, the Tychonoff Plank is the product of the successor of the first uncountable ordinal and the successor of the first infinite ordinal. The deleted Tychonoff plank is $X=[0,\omega_1] \times [0,\omega]-\lbrace{(\omega_1,\omega)}\rbrace$, i.e., the Tychonoff plank minus the corner point $(\omega_1,\omega)$. This is another famous example of topological spaces that often appear in first year topology courses. My goal here is to present two facts. The deleted Tychonoff plank is not normal, thus showing that the Tychonoff plank $X=[0,\omega_1] \times [0,\omega]$ not hereditarily normal. The second point is that for the deleted Tychonoff plank, the one-point compactification and the Stone-Cech compactification conincide. This is due to the fact that any continuous real-valued function defined on the deleted Tychonoff plank can be extended to a continuous function on the product space $X=[0,\omega_1] \times [0,\omega]$ (see Theorem 19.12 in [Willard]).

First we show the deleted Tychonoff plank $X$ is not normal. Let $H=\lbrace{(\alpha,\omega):\alpha<\omega_1}\rbrace$ and $K=\lbrace{(\omega_1,n):n<\omega}\rbrace$. These are two disjoint closed sets in $X$. Let $H \subset U$ and $K \subset V$ where $U$ and $V$ are open in $X$. I will show $U \cap V \neq \phi$.

First define a pressing down function on $H$. For each limit ordinal $\alpha<\omega_1$, there is an integer $m(\alpha)$ and there is a countable ordinal $g(\alpha)<\alpha$ such that $[g(\alpha),\alpha] \times [m(\alpha),\omega] \subset U$. By the pressing down lemma, there is $\delta<\omega_1$ and there is a stationary set $S$ such that all points in $S$ are mapped to $\delta$ by $g$. Choose an uncountable $S_0 \subset S$ such that for each $\alpha \in S_0$, $m(\alpha)=j$ for some $j$. Consider $K_0=\lbrace{(\omega_1,n):j \leq n<\omega}\rbrace \subset K$.

Claim. $K_0 \subset Cl(U)$ where $Cl$ is the closure operator. Fix $x=(\omega_1,n)$. Fix an open set $O$ containing $x$, say $O=[\gamma,\omega_1] \times \lbrace{n}\rbrace$ where $\delta<\gamma<\omega_1$. Choose $\alpha \in S_0$ such that $\delta<\gamma<\alpha$. Based on how $g$ is obtained, $[\delta,\alpha] \times [j,\omega] \subset U$. It follows that $[\gamma,\alpha] \times \lbrace{n}\rbrace \subset U$ and $[\gamma,\alpha] \subset [\gamma,\omega_1] \times \lbrace{n}\rbrace=O$. So any open set $O$ containing $x$ contains points of $U$. Thus the claim is established.

Since $K_0 \subset K \subset V$, $V$ would contain points of $U$. Thus the deleted Tychonoff plank $X$ is not normal.

To established the second point, let $f:X\rightarrow\mathbb{R}$ a continuous function. Then $f$ can be  extended to include the corner point. Consider the restriction of $f$ to the following horizontal segments of the deleted Tychonoff plank.
$A=\lbrace{(\alpha,\omega):\alpha<\omega_1}\rbrace$
$B_n=\lbrace{(\alpha,n):\alpha \leq \omega_1}\rbrace$ for each $n<\omega$.

Any continuous function is eventually constant on these horizontal segments. For the proof, see this post. So there is $\alpha_0<\omega_1$ such that $f((\alpha,\omega))=x_0$ for all $(\alpha,\omega) \in [\alpha_0,\omega_1) \times \lbrace{\omega}\rbrace$. For each $n<\omega$, there is $\beta_n<\omega_1$ such that $f((\alpha,n))=y_n$ for all $(\alpha,n) \in [\beta_n,\omega_1] \times \lbrace{n}\rbrace$. Choose some $\alpha>\alpha_0,\beta_n$ for all $n$.

It is clear that the sequence $\lbrace{y_n}\rbrace$ converges to $x_0$. So by defining $f((\omega_1,\omega))=x_0$, $f$ is still a continuous function.

Reference
[Willard] Willard, S., [1970] General Topology, Addison-Wesley Publishing Company, Inc.

# Normal x Compact Needs Not be Normal

This is to present a non-normal cross product where one factor is a normal space and the other factor is a compact space. The normal space is $X=[0,\omega_1)$ and the compact space is $Y=[0,\omega_1]$. I would like to show that $X \times Y$ is not normal. For a basic discussion on these ordinal spaces with the order topology, see this previous post.

In addition to demonstrating an example meeting the criteria set forth in the title, I would like to look at this through two angles. Because these spaces have the ordered topology, they are hereditarily normal (completely normal). Note that $X \times Y$ is a subspace of $Y \times Y$. Thus $Y \times Y$ fails to be hereditarily normal, demonstrating that hereditarily normal x hereditarily normal needs not be hereditarily normal. Another example of this is the double arrow space. In this previous post, I showed that the square of the double arrow space is not hereditarily normal. The double arrow space is perfectly normal and is thus a stronger example in this regard since it shows the square of a perfectly normal space needs not be hereditarily normal. But the uncountable ordinal spaces of $[0,\omega_1)$ and $[0,\omega_1]$ are important counterexamples as well as building blocks for other counterexamples. So for the record, I present them here.

For the second angle, these two spaces show that hereditarily normality is not “nice” enough to prevent non-normal product.

In proving $X \times Y$ is not normal, the Pressing Down Lemma is used. A subset $S$ of $X=[0,\omega_1)$ is a stationary set if $S$ meets every closed and unbounded set in $X$. The following is one version of the Pressing Down Lemma (see Lemma 6.15 on p. 80 of [1] for a more general version).

Pressing Down Lemma

Let $S$ be a stationary subset of $[0,\omega_1)$. Let $f:S \rightarrow \omega_1$ such that for each $\gamma \in S$, $f(\gamma)<\gamma$, then for some $\alpha<\omega_1$, $f^{-1}\lbrace{\alpha}\rbrace$ is a stationary subset of $[0,\omega_1)$.

To show that $X \times Y$ is not normal, let $H$ and $K$ be the following sets.

$H=\lbrace{(\alpha,\alpha):\alpha<\omega_1}\rbrace$

$K=\lbrace{(\alpha,\omega_1):\alpha<\omega_1}\rbrace$

These two sets are disjoint and closed in $X \times Y$. Let $U$ and $V$ be open such that $H \subset U$ and $K \subset V$. It follows that $U \cap V \neq \phi$.

For each $\alpha<\omega_1$, let $g(\alpha)<\omega_1$ such that $[g(\alpha),\alpha] \times [g(\alpha),\alpha] \subset U$. Since $g$ is a pressing down function, there is some $\delta<\omega_1$ and there is a stationary set $S$ such that all points in $S$ are mapped to $\delta$ by $g$. Specifically, for each $\alpha \in S$, we have:

$[\delta,\alpha] \times [\delta,\alpha] \subset U$

Choose $\beta_0>\delta$ and let $\beta=\beta_0+1$. We have $(\beta,\omega_1) \in K \subset V$. Choose $\gamma>\delta$ such that

$\lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V$

Choose $\alpha \in S$ such that $\delta < \beta<\alpha$ and $\delta<\gamma<\alpha<\omega_1$. We have the following set inclusions:

$\lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset [\delta,\alpha] \times [\delta,\alpha] \subset U$

$\lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset \lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V$

Thus $U \cap V \neq \phi$. This completes the proof that the product $X \times Y$ is not normal.

____________________________________________________________________

Reference

1. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

____________________________________________________________________

Revised February 3, 2014.
$\copyright \ 2014 \text{ by Dan Ma}$

# CCC + Paracompact => Lindelof

It is a well known fact in general topology that in the class of spaces with the countable chain condition (CCC), paracompactness equals the Lindelof property. Any regular Lindelof space is paracompact. This post gives a proof that any paracompact space with the CCC is a Lindelof space.

A space has the countable chain condition (CCC) if every pairwise disjoint family of open sets in the space is countable. Clearly any separable space has the CCC. Other examples of spaces with the CCC are product of separable spaces and function spaces $C_p(X)$ where $X$ is any completely regular space.

CCC + Paracompact implies Lindelof

Suppose that the space $X$ is a paracompact with the CCC. Let $\mathcal{U}$ be an open cover of $X$. By the paracompactness of $X$, there is $\mathcal{U}_0$, which is a locally finite open refinement of $\mathcal{U}$.

For each $x \in X$, choose $V_x \in \mathcal{U}_0$ such that $x \in V_x$ and choose $O_x$, an open subset of $X$ such that $x \in O_x$ and $O_x$ meets only finitely many elements of $\mathcal{U}_0$. For each $x \in X$, let $C_x=O_x \cap V_x$. Let $\mathcal{C}=\{ C_x: x \in X \}$. One observation about $\mathcal{C}$ is that it is an open cover of $X$. Another observation about $\mathcal{C}$ is that each element of $\mathcal{C}$ meets only finitely many elements of $\mathcal{U}_0$.

For any $E,F \in \mathcal{C}$, the finite set $\{G_1,G_2,\cdots,G_n \} \subset \mathcal{C}$ is said to be a chain from $E$ to $F$ if $E=G_1$, $F=G_n$ and $G_i \cap G_{i+1} \ne \varnothing$ for all $1 \le i. For each $E \in \mathcal{C}$, define $T(E)$ and $S(E)$ as follows:

$T(E)=\{ F \in \mathcal{C}: \exists \text{ a chain from } E \text{ to } F \}$

$S(E)=\bigcup T(E)$

Observe that each $T(E)$ is a countable subcollection of $\mathcal{C}$. For $E_1,E_2 \in \mathcal{C}$, if $S(E_1) \cap S(E_2) \ne \varnothing$, then there is a chain from $E_1$ to $E_2$. It follows that $T(E_1)=T(E_2)$ and $S(E_1)=S(E_2)$. Let $\mathcal{W}$ be the family of all distinct $S(E)$ where $E \in \mathcal{C}$. By the CCC of the space $X$, $\mathcal{W}$ must be countable.

Note that $\mathcal{W}$ is also an open cover of $X$. As observed previously, for each $E \in \mathcal{C}$, $T(E)$ is countable. For each $F \in T(E)$, $F \subset V$ for some $V \in \mathcal{U}_0$. Thus each $S(E) \in \mathcal{W}$ is matched with countably many $V \in \mathcal{U}_0$. Since $\mathcal{W}$ is countable, there are countably many $V \in \mathcal{U}_0$ that associate with the elements of $\mathcal{W}$. Since $\mathcal{W}$ is a cover of $X$, the associated countably many $V \in \mathcal{U}_0$ must form an open cover of $X$ too. It follows that countably many elements of the original cover $\mathcal{U}$ form a cover of $X$. This concludes the proof that $X$ is Lindelof.

$\text{ }$

$\text{ }$

$\text{ }$

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2009 – 2019 – Dan Ma

Revised June 15, 2019