# The Normality of Products Under Perfect Preimages

Lucas D. O’Brien

###### Abstract.

A proof of the following theorem is given, answering an open problem
attributed to Kunen: suppose that $T$ is compact and that $Y$ is the image of
$X$ under a perfect map, $X$ is normal, and $Y\times T$ is normal. Then
$X\times T$ is normal.

††2020 Mathematics Subject Classification. Primary 54B10, 54C10, 54D15††Key
words and phrases. Normality of products, perfect map, compact.

## 1\. Introduction

In [Rudin73], Mary Ellen Rudin proved that given a compact space $T$, a normal
space $X$, and a space $Y$ that is the image of $X$ under a closed map, that
$X\times T$ is normal only if $Y\times T$ is normal. Rudin’s proof relied on a
characterization of normality of products with a compact factor which appeared
as Theorem 3 in [Rudin73], and is included as Theorem 2.7 in this paper. This
characterization loosely says that the normality of the product $X\times T$ is
equivalent to certain open covers of $X$ having locally finite open
refinements. Rudin called these open covers $X$-separations; we follow the
convention of [Kunen84] and call them $\mathscr{B}$-coverings (see Definition
2.3). In view of Theorem 2.7, there is correlation between paracompactness and
the normality of a product with a compact factor. Given a compact space $T$,
Morita defines a space $X$ to be $T$-paracompact provided $X\times T$ is
normal [Morita61].

We may intuitively expect the property of $T$-paracompactness to be preserved
in a similar way to paracompactness under mappings. Indeed, Rudin’s theorem
shows that $T$-paracompactness is preserved under images of closed maps. The
question of whether $T$-paracompactness is preserved under preimages of
perfect maps is attributed to Kunen and appeared as an open problem in
[Kunen84] and [Rudin75]; this paper shows that this is indeed the case.

By the Tietze extension theorem, any continuous map from a closed subspace of
a normal space $X$ to $\mathbb{R}$ can be extended to a continuous map of all
of $X$ into $\mathbb{R}$. Therefore, the study of the normality of products
has applications to the problem of the extendability of continuous functions
from subsets of product spaces. For more details, see [Reed80].

## 2\. Characterization of Normality in Products with a Compact Factor

In the following, $T$ denotes a compact Hausdorff space and $\mathscr{B}$
denotes a base of $T$ which is closed with respect to finite unions and
intersections. A normal space refers to a topological space which is both
$\text{T}_{4}$ and $\text{T}_{2}$.

######  Definition 2.1.

Let $\mathcal{A}$ be a collection of sets, and let $\mathcal{B}$ be a
collection of ordered pairs of sets. We will say a map
$G:\mathcal{B}\to\mathcal{A}$ is intersection preserving provided $G(A,B)\cap
G(A^{\prime},B^{\prime})=G(A\cap A^{\prime},B\cap B^{\prime})$ for all
$(A,B),(A^{\prime},B^{\prime})\in\mathcal{B}$.

######  Remark 2.2.

Suppose $G:\mathcal{B}\to\mathcal{A}$ is intersection preserving. Then, for
$(A,B),(A^{\prime},B^{\prime})\in\mathcal{B}$ such that $A\subseteq
A^{\prime}$ and $B\subseteq B^{\prime}$, we have $G(A,B)\subseteq
G(A^{\prime},B^{\prime})$.

######  Definition 2.3.

[Kunen84] Fix a base $\mathscr{B}$ of $T$, and define
$\mathcal{S}_{\mathscr{B}}=\\{(B,D)\in\mathscr{B}^{2}|\
\overline{B}\cap\overline{D}=\emptyset\\}$. An open cover $\mathscr{G}$ of a
space $X$ is called a $\mathscr{B}$-covering of $X$ if there exists an
intersection preserving surjection
$G:\mathcal{S}_{\mathscr{B}}\to\mathscr{G}$. We call $G$ the associated map to
the $\mathscr{B}$-covering.

Let $X$ be a space, and let $G:\mathcal{S}_{\mathscr{B}}\to\mathscr{P}(X)$ be
an intersection preserving map. Then, the set $\mathscr{G}=\\{G(B,D)\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}$ is a $\mathscr{B}$-covering of $X$ if
and only if $\mathscr{G}$ is an open cover of $X$.

######  Definition 2.4.

An open cover $\mathscr{G}^{\prime}$ of a space $X$ is called a
pre-$\mathscr{B}$-covering of $X$ if there exists an intersection preserving
surjection $G:\mathscr{B}^{2}\to\mathscr{G}^{\prime}$ and
$\mathscr{G}=\\{G(B,D)\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}\subseteq\mathscr{G}^{\prime}$ is a
$\mathscr{B}$-covering of $X$. We call $\mathscr{G}$ the
$\mathscr{B}$-covering associated with $\mathscr{G}^{\prime}$.

######  Example 2.5.

Let $X$ be a normal space, and let $K,L\subseteq X\times T$ be closed,
disjoint sets. For $B,D\in\mathscr{B}$, define

$G(B,D)=\\{x\in X\ |\ K_{x}\subseteq B\text{ and }L_{x}\subseteq D\\},$

where $K_{x}=\\{t\in T\ |\ (x,t)\in K\\}$ and $L_{x}=\\{t\in T\ |\ (x,t)\in
L\\}$. Then, the set $\mathscr{G}^{\prime}=\\{G(B,D)\ |\
(B,D)\in\mathscr{B}^{2}\\}$ is a pre-$\mathscr{B}$-covering of $X$.

Indeed, by definition, $G:\mathscr{B}^{2}\to\mathscr{G},(B,D)\mapsto G(B,D)$
is surjective, and it is easy to check that $G$ is intersection preserving.
Thus, it remains to show that $\mathscr{G}=\\{G(B,D)\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}$ covers $X$ and $G(B,D)$ is open for each
$(B,D)\in\mathscr{B}^{2}$. Let $x\in X$. Since $K$ and $L$ are disjoint, so
are $K_{x}$ and $L_{x}$. Furthermore, notice that
$K_{x}=\pi_{T}(K\cap(\\{x\\}\times T))$ and $L_{x}=\pi_{T}(L\cap(\\{x\\}\times
T))$, where $\pi_{T}:X\times T\to T$ is the projection map. Since
$K\cap(\\{x\\}\times T)$ and $L\cap(\\{x\\}\times T)$ are compact, $K_{x}$ and
$L_{x}$ are closed. Thus, there exist $B,D\in\mathscr{B}$ such that
$K_{x}\subseteq B$, $L_{x}\subseteq D$ and
$\overline{B}\cap\overline{D}=\emptyset$ since $\mathscr{B}$ is closed with
respect to finite unions. Then, $x\in G(B,D)$. Notice that for each
$(B,D)\in\mathscr{B}^{2}$, we may write

$G(B,D)=X\setminus\pi_{X}((K\setminus(X\times B))\cup(L\setminus(X\times
D))),$

where $\pi_{X}:X\times T\to X$ is the projection map. Since $T$ is compact,
$\pi_{X}$ is a closed map, and therefore $G(B,D)$ is open.

The next lemma follows from the techniques used in [Kunen84].

######  Lemma 2.6.

Let $X$ be a normal space. For every $\mathscr{B}$-covering $\mathscr{G}$ of
$X$, there exists a pre-$\mathscr{B}$-covering $\mathscr{H}^{\prime}$ of $X$
such that the $\mathscr{B}$-covering associated with $\mathscr{H}^{\prime}$ is
an open cover refinement of $\mathscr{G}$.

###### Proof.

Let $\mathscr{G}$ be a $\mathscr{B}$-covering of $X$, and let
$G:\mathcal{S}_{\mathscr{B}}\to\mathscr{G}$ be the associated map. Define

$K=(X\times T)\setminus\bigcup\\{G(B,D)\times(T\setminus\overline{B})\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}$

and

$L=(X\times T)\setminus\bigcup\\{G(B,D)\times(T\setminus\overline{D})\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}.$

Evidently $K$ and $L$ are closed, we will show they are disjoint. Indeed, let
$(x,t)\in X\times T$. Since $\mathscr{G}$ covers $X$, there exists
$(B,D)\in\mathcal{S}_{\mathscr{B}}$ such that $x\in G(B,D)$. We must have
$t\notin\overline{B}$ or $t\notin\overline{D}$, and therefore either $(x,t)\in
G(B,D)\times(T\setminus\overline{B})$ or $(x,t)\in
G(B,D)\times(T\setminus\overline{D})$, so by the definition of $K$ and $L$, we
must have either $(x,t)\in(X\times T)\setminus K$ or $(x,t)\in(X\times
T)\setminus L$. Therefore, $K$ and $L$ are disjoint.

For $B,D\in\mathscr{B}$, define

$H(B,D)=\\{x\in X\ |\ K_{x}\subseteq B\text{ and }L_{x}\subseteq D\\},$

where $K_{x}=\\{t\in T\ |\ (x,t)\in K\\}$ and $L_{x}=\\{t\in T|(x,t)\in L\\}$.
As in Example 2.5, $\mathscr{H}^{\prime}=\\{H(B,D)\ |\
(B,D)\in\mathscr{B}^{2}\\}$ is a pre-$\mathscr{B}$-covering of $X$. Therefore,
to show that the $\mathscr{B}$-covering associated with $\mathscr{H}^{\prime}$
refines $\mathscr{G}$, it remains to show that $H(B,D)\subseteq G(B,D)$ for
all $(B,D)\in\mathcal{S}_{\mathscr{B}}$.

Let $(B,D)\in\mathcal{S}_{\mathscr{B}}$, and let $x\in H(B,D)$. Then,
$K_{x}\subseteq B$ and $L_{x}\subseteq D$. By definition of $K$ and $L$, we
have

$\\{x\\}\times(T\setminus
B)\subseteq\bigcup\\{G(B^{\prime},D^{\prime})\times(T\setminus\overline{B^{\prime}})\
|\
B^{\prime},D^{\prime}\in\mathscr{B},\overline{B^{\prime}}\cap\overline{D^{\prime}}=\emptyset\\},$

and

$\\{x\\}\times(T\setminus
D)\subseteq\bigcup\\{G(B^{\prime\prime},D^{\prime\prime})\times(T\setminus\overline{D^{\prime\prime}})\
|\
B^{\prime\prime},D^{\prime\prime}\in\mathscr{B},\overline{B^{\prime\prime}}\cap\overline{D^{\prime\prime}}=\emptyset\\}.$

Since $\\{x\\}\times(T\setminus B)$ is compact, it is covered by finitely many
elements
$\\{G(B^{\prime}_{i},D^{\prime}_{i})\times(T\setminus\overline{B_{i}^{\prime}})|1\leq
i\leq n\\}$. Therefore, we have
$T-B\subseteq\bigcup_{i=1}^{n}(T\setminus\overline{B_{i}^{\prime}})$ and
$x\in\bigcap_{i=1}^{n}G(B_{i},D_{i})$. Similarly, we may find a finite family
$\\{(B^{\prime\prime}_{1},D^{\prime\prime}_{1}),\dots,(B^{\prime\prime}_{m},D^{\prime\prime}_{m})\\}$
such that
$T-D\subseteq\bigcup_{i=1}^{m}(T\setminus\overline{D_{i}^{\prime\prime}})$ and
$x\in\bigcap_{i=1}^{m}G(B^{\prime\prime}_{i},D^{\prime\prime}_{i})$.
Therefore, we have $B\subseteq\bigcap_{i=1}^{n}B^{\prime}_{i}$,
$D\subseteq\bigcap_{i=1}^{m}D^{\prime\prime}_{i}$, and

$x\in
G(\bigcap_{i=1}^{n}B^{\prime}_{i}\cap\bigcap_{i=1}^{m}B^{\prime\prime}_{i},\bigcap_{i=1}^{n}D^{\prime}_{i}\cap\bigcap_{i=1}^{m}D^{\prime\prime}_{i})\subseteq
G(B,D).$

Thus, we have that $H(B,D)\subseteq G(B,D)$ for all
$(B,D)\in\mathcal{S}_{\mathscr{B}}$. Therefore, the $\mathscr{B}$-covering
associated with $\mathscr{H}^{\prime}$ refines $\mathscr{G}$. ∎

The proof of Theorem 3.1 relies on the following characterization of normality
of products with a compact factor, which allows us to work with
$\mathscr{B}$-coverings instead of disjoint closed sets. The proof of this
particular statement of Rudin’s Lemma can be found in [Kunen84].

######  Theorem 2.7 (Rudin’s Lemma).

[Kunen84, Rudin73, Beslagic91] The product space $X\times T$ is normal if and
only if $X$ is normal and every $\mathscr{B}$-covering of $X$ has a locally
finite open refinement.

## 3\. Proof of Theorem

######  Theorem 3.1.

Suppose that $T$ is compact and that $Y$ is the image of $X$ under a perfect
map $p$, $X$ is normal, and $Y\times T$ is normal. Then, $X\times T$ is
normal.

###### Proof.

We may assume without loss of generality that $T$ is not a singleton.

By Rudin’s Lemma, it suffices to show that every $\mathscr{B}$-covering of $X$
has a locally finite open refinement. For any $\mathscr{B}$-covering
$\mathscr{G}$ of $X$, there exists a pre-$\mathscr{B}$-covering whose
associated $\mathscr{B}$-covering refines $\mathscr{G}$ by Lemma 2.6, so it is
enough to show that for any pre-$\mathscr{B}$-covering of $X$, the associated
$\mathscr{B}$-covering has a locally finite open refinement.

Let $\mathscr{G}^{\prime}$ be a pre-$\mathscr{B}$-covering of $X$, and let
$G:\mathscr{B}^{2}\to\mathscr{G}^{\prime}$ be the map associated with
$\mathscr{G}^{\prime}$. We will show that
$\mathscr{G}=\\{G(B,D)|(B,D)\in\mathcal{S}_{\mathscr{B}}\\}$ has a locally
finite open refinement. For each $(B,D)\in\mathscr{B}^{2}$, define

$H(B,D)=\\{y\in Y\ |\ p^{-1}\\{y\\}\subseteq G(B,D)\\}.$

It is easy to check that $H$ is intersection preserving. Define
${\mathscr{H}^{\prime}=\\{H(B,D)|(B,D)\in\mathscr{B}^{2}\\}}$; we claim that
$\mathscr{H}^{\prime}$ is an open cover of $Y$. Indeed, for each $y\in Y$,
$p^{-1}\\{y\\}$ can be covered by finitely many elements
$\\{G(B_{1},D_{1}),\dots,G(B_{n},D_{n})\\}$ of $\mathscr{G}$. Then, since
$\mathscr{B}$ is closed under finite union, by Remark 2.2,

$p^{-1}\\{y\\}\subseteq\bigcup_{i=1}^{n}G(B_{i},D_{i})\subseteq
G(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})\in\mathscr{G}^{\prime},$

so $y\in H(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})$. Furthermore, since
$G(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})$ is an open set containing
$p^{-1}\\{y\\}$, from Theorem 1.4.13 in [Engelking89] we know there is some
neighbourhood $U$ of $y$ such that $p^{-1}(U)\subseteq
G(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})$; therefore $U\subseteq
H(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})$ and so
$H(\bigcup_{i=1}^{n}B_{i},\bigcup_{i=1}^{n}D_{i})$ is open.

For each $B,D\in\mathscr{B}$, define

$K_{B}=((Y\setminus H(B,T))\times T)\cup(Y\times\overline{B})$

and

$L_{D}=((Y\setminus H(T,D))\times T)\cup(Y\times\overline{D}).$

Define

$K=\bigcap_{B\in\mathscr{B}}K_{B},\ \ \ \ \ \ \ \
L=\bigcap_{D\in\mathscr{B}}L_{D}.$

Evidently $K$ and $L$ are closed. We show they are disjoint; indeed, note that
for $B,D\in\mathscr{B}$, we have

$K_{B}\cap L_{D}=((Y\setminus(H(B,T)\cup H(T,D)))\times T)\cup((Y\setminus
H(T,D))\times\overline{B})\cup((Y\setminus
H(B,T))\times\overline{D})\cup(Y\times(\overline{B}\cap\overline{D})).$

It suffices to show that each of the terms in the above union are individually
empty when intersected over all pairs $(B,D)\in\mathscr{B}^{2}$. The second
and third terms are both empty when this intersection is taken since
$\mathscr{H}^{\prime}$ covers $Y$, and for any
$H(B,D)\in\mathscr{H}^{\prime}$, we have $H(B,D)\subseteq H(B,T)$ and
$H(B,D)\subseteq H(T,D)$. Since $T$ is normal and contains at least two
distinct points, there are basis elements $B,D\in\mathscr{B}$ such that
$\overline{B}\cap\overline{D}=\emptyset$, so the fourth term is empty under
this intersection. Thus, we have

$K\cap L=\bigcap_{(B,D)\in\mathscr{B}^{2}}K_{B}\cap
L_{D}=((Y\setminus\bigcup_{(B,D)\in\mathscr{B}^{2}}(H(B,T)\cup H(T,D)))\times
T).$

Since $\mathscr{H}^{\prime}$ covers $Y$, we have

$\bigcup_{(B,D)\in\mathscr{B}^{2}}H(B,T)\cup
H(T,D)\supseteq\bigcup_{(B,D)\in\mathscr{B}^{2}}H(B,T)\cap
H(T,D)=\bigcup_{(B,D)\in\mathscr{B}^{2}}H(B,D)=Y,$

and therefore $K\cap L=\emptyset$.

We claim that $\mathscr{H}=\\{H(B,D)\ |\
(B,D)\in\mathcal{S}_{\mathscr{B}}\\}\subseteq\mathscr{H}^{\prime}$ is a
$\mathscr{B}$-covering of $Y$. It remains to show that $\mathscr{H}$ covers
$Y$. Let $y\in Y$. As shown in Example 2.5 $K_{y}=\\{t\in T\ |\ (y,t)\in K\\}$
and $L_{y}=\\{t\in T\ |\ (y,t)\in L\\}$ are closed disjoint subsets of $T$, so
there exist $B,D\in\mathscr{B}$ with $\overline{B}\cap\overline{D}=\emptyset$
such that $K_{y}\subseteq B$ and $L_{y}\subseteq D$ by the fact that
$\mathscr{B}$ is closed under finite unions. We show that $y\in H(B,T)$, an
analogous argument shows that $y\in H(T,D)$ and so $y\in H(B,T)\cap
H(T,D)=H(B,D)$.

Define $\mathcal{J}=\\{B^{\prime}\in\mathscr{B}\ |\ y\in H(B^{\prime},T)\\}$,
we claim that there is some $B^{\prime}\in\mathcal{J}$ such that
$B^{\prime}\subseteq B$, and thus $y\in H(B^{\prime},T)\subseteq H(B,T)$. By
definition of $K$, we must have
$\bigcap_{B^{\prime}\in\mathcal{J}}\overline{B^{\prime}}\subseteq
K_{y}\subseteq B$. Therefore, $\\{T\setminus\overline{B^{\prime}}\ |\
B^{\prime}\in\mathcal{J}\\}$ is an open cover of $T\setminus B$, so there
exists a finite subcover
$\\{T\setminus\overline{B^{\prime}_{1}},\dots,T\setminus\overline{B^{\prime}_{n}}\\}$.
In particular, we have $\bigcap_{i=1}^{n}\overline{B^{\prime}_{i}}\subseteq
B$. But $y\in H(B^{\prime}_{i},T)$ for each $1\leq i\leq n$, and therefore

$y\in\bigcap_{i=1}^{n}H(B^{\prime}_{i},T)\subseteq
H(\bigcap_{i=1}^{n}B^{\prime}_{i},T)$

so $B^{\prime\prime}=\bigcap_{i=1}^{n}B^{\prime}_{i}\in\mathcal{J}$. But since
$B^{\prime\prime}\subseteq B$, we have

$y\in H(B^{\prime\prime},T)\subseteq H(B,T),$

and so $y\in H(B,T)$. Therefore, for each $y\in Y$, there exist
$(B,D)\in\mathcal{S}_{\mathscr{B}}$ such that $y\in H(B,D)$.

Since $\mathscr{H}$ is a $\mathscr{B}$-covering of $Y$, by Rudin’s Lemma, it
has a locally finite open refinement $\mathscr{V}$. For each
$V\in\mathscr{V}$, there exists $(B_{V},D_{V})\in\mathcal{S}_{\mathscr{B}}$
such that $V\subseteq H(B_{V},D_{V})$, so by definition of
$\mathscr{H}^{\prime}$, we have that for each $V\in\mathscr{V}$,

$p^{-1}(V)\subseteq p^{-1}(H(B_{V},D_{V}))\subseteq G(B_{V},D_{V}).$

Therefore, the set $\\{p^{-1}(V)\ |\ V\in\mathscr{V}\\}$ is a locally finite
open refinement of $\mathscr{G}$. Thus, by Rudin’s Lemma, $X\times T$ is
normal.

∎

This allows us to answer another problem in [Kunen84].

######  Corollary 3.2.

Suppose $T$ is compact, and $X\times T$ is normal. Then, $X\times T^{n}$ is
normal for each $n\in\mathbb{N}$.

###### Proof.

This follows by induction on $n$, noticing that the projection $p:X\times T\to
X$ is perfect because $T$ is compact. ∎

Corollary 3.2 does not hold when $n$ is replaced with $\omega$. To see why, we
need the following lemma.

######  Lemma 3.3.

[Morita61] Let $Q$ and $Q^{\prime}$ be any two compact Hausdorff spaces. If
$Q^{\prime}$ is either a closed subset or a continuous image of $Q$, then if
$X\times Q$ is normal, so is $X\times Q^{\prime}$.

######  Proposition 3.4.

There exists a space $X$ and a compact set $T$ such that $X\times T$ is normal
but $X\times T^{\omega}$ is not normal.

###### Proof.

Let $T=2$ with the discrete topology, and let $X$ be a Dowker space, that is,
a normal space for which $X\times[0,1]$ is not normal [Rudin71]. Since $T$ is
discrete and $X$ is normal, we know that $X\times T$ is normal. Suppose that
$X\times 2^{\omega}$ is normal. Since $2^{\omega}$ is the Cantor set, $[0,1]$
is the continuous image of $2^{\omega}$ (see [Engelking89]), so by Lemma 3.3,
$X\times[0,1]$ is normal, contradicting the fact that $X$ is Dowker. ∎

## Acknowledgements

The author would like to thank Clovis Hamel and Franklin D. Tall for their
invaluable comments, corrections, and references.

## References