Munkres Topology Solution: Chapter 6

§39. Local Finiteness

Ex.39.1

Check the statements in Example 1.

Omitted.

Ex.39.2

Find a point-finite open covering $\mathscr{A}$ of $\Bbb R$ that is not locally finite. (The collection $\mathscr{A}$ is point-finite if each point of $\Bbb R$ lies in only finitely many elements of $\mathscr{A}$.)

$\lbrace\Bbb R\rbrace\cup\lbrace(1/(n+1),1/n)\rbrace_n$

Ex.39.3

Give an example of a collection of sets $\mathscr{A}$ that is not locally finite, such that the collection $\mathscr{B}=\lbrace\overline A|A\in\mathscr{A}\rbrace$ is locally finite.

Note that if different $A\in\mathscr{A}$ give different closure then the collection $\mathscr{B}$ is definitely not locally finite. So the counterexample happens when closures of many $A\in\mathscr{A}$ coincide.

For example, $\lbrace A\subset\Bbb R|A\supset\Bbb Q\rbrace$ will do.

Ex.39.4

Let $\mathscr{A}$ be the following collection of subsets of $\Bbb R$:
$$\mathscr{A}=\lbrace(n,n+2)|n\in\Bbb Z\rbrace$$ Which of the following collections refine $\mathscr{A}$?
$$
\begin{cases}
\mathscr{B}=\lbrace(x,x+1)|x\in\Bbb R\rbrace\\
\mathscr{C}=\lbrace(n,n+3/2)|n\in\Bbb Z\rbrace\\
\mathscr{D}=\lbrace(x,x+3/2)|x\in\Bbb R\rbrace
\end{cases}
$$

$\mathscr{B},\mathscr{C}$. Note that $(0.75,2.25)$ lies in $\mathscr{D}$ but not in $\mathscr{A}$.

Ex.39.5

Show that if $X$ has a countable basis, a collection $\mathscr{A}$ of subsets of $X$ is countably locally finite if and only if it is countable.

If $\mathscr{A }$ is countable, then it is of course countably locally finite.

Conversely, if $\mathscr{A }$ is countably locally finite, say $\mathscr{A }=\bigcup\mathscr{A}_n$. Let $\lbrace B_m\rbrace$ be a countable basis for $X$. For each $n\in\Bbb Z,x\in X$, there is an open nbhd $U_x$ of $x$ such that $U_x$ intersects finitely many elements of $\mathscr{A}_n$. Choose an $B_m$ such that $x\in B_m\subset U_x$, then $B_m$ intersects finitely many elements of $\mathscr{A}_n$. Therefore, we see that each $B_m$ intersects finitely many elements of $\mathscr{A}_n$. Hence there are only countably many elements in $\mathscr{A}_n$. Thus $\mathscr{A}$ is countable.

Ex.39.6

Consider $\Bbb R^\omega$ in the uniform topology. Given $n$, let $\mathscr{B}_n$ be the collection of all subsets of $\Bbb R^\omega$ of the form $\prod A_i$ , where $A_i=\Bbb R$ for $i\le n$ and $A_i$ equals either $\lbrace 0\rbrace$ or $\lbrace 1\rbrace$ otherwise. Show that the collection $\mathscr{B}=\bigcup_n\mathscr{B}_n$ is countably locally finite, but neither countable nor locally finite.

First, it is easy to see $\mathscr{A}_n$ is even locally discrete: for each open ball $B(x,1/3)$, if it intersects two subsets of the form $\prod A_i\in\mathscr{B}_n$, then we see $\mathrm{diam}B(x,1/3)\ge 1$, a contradiction.

Second, $|\mathscr{B}|=\sum 2^{\aleph_0}=\aleph_0 2^{\aleph_0}=\mathfrak c$, uncountable.

Finally, pick $0\in\Bbb R^\omega$, then $0\in\Bbb R^{n}\times\prod_{i\ge n+1}\lbrace 0\rbrace\in\mathscr{B}_n$ for each $n\in\Bbb Z^+$, thus any open nbhd of $0$ will intersect infinitely many elements in $\mathscr{B}$

§40. The Nagata-Smirnov Metrization Theorem

Ex.40.1

Check the details of Examples 1 and 2.

Omitted.

Ex.40.2

A subset $W$ of $X$ is said to be an “$F_\sigma$ set” in $X$ if $W$ equals a countable union of closed sets of $X$. Show that $W$ is an $F_\sigma$ set in $X$ if and only if $X-W$ is a $G_\delta$ set in $X$.
[The terminology comes from the French. The “F” stands for “fermé,” which means “closed,” and the “$\sigma$” for “somme,” which means “union.”]

Omitted.

Ex.40.3

Many spaces have countable bases; but no $T_1$ space has a locally finite basis unless it is discrete. Prove this fact.

Suppose $X$ is a $T_1$ space with a locally finite basis. We know that in a $T_1$ space, the intersection of all nbhd of a point $\bigcap\mathcal N(x)$ equals to the point itself. Let $\mathscr{B}$ be a locally finite basis for $X$, $\mathscr{C}$ a subcollection of it containing all basis element containing $x$. By local finiteness, there exists an open nbhd $U$ of $x$ such that $U$ intersects finitely many elements of $\mathscr{C}$, say $C_1,…,C_n$. Then $\lbrace x\rbrace=\bigcap\mathcal N(x)=U\cap C_1\cap…\cap C_n$ is an open set, proving the discreteness of $X$.

Ex.40.4

Find a non-discrete space that has a countably locally finite basis but does not have a countable basis.

By Lemma 39.2, every metrizable space has a countably locally finite basis. So it remains us to find a non-discrete, non-$C_2$ metrizable space. For example $\Bbb R^\omega$ in the uniform topology.

Ex.40.5

A collection $\mathscr{A}$ of subsets of $X$ is said to be locally discrete if each point of $X$ has a neighborhood that intersects at most one element of $\mathscr{A}$. A collection $\mathscr{B}$ is countably locally discrete (or “$\sigma$-locally discrete”) if it equals a countable union of locally discrete collections. Prove the following:
Theorem (Bing metrization theorem). A space $X$ is metrizable if and only if it is regular and has a basis that is countably locally discrete.

I believe the proof is just a copy of Theorem 40.3. The only thing we need to do is converting the dependent lemmas into locally discrete version. I will list them below, the proof of which is exactly the same as their original version, changing some words.

Lemma. Let $\mathscr{A}$ be a collection of subsets of $X$ that is locally discrete.
(1) Any subcollection of $\mathscr{A}$ is locally discrete.
(2) The collection $\mathscr{B}=\lbrace \overline A|A\in\mathscr{A}\rbrace$ is locally discrete.
(3) $\overline{\bigcup_{A\in\mathscr{A}}A}=\bigcup_{A\in\mathscr{A}}\overline A$

Lemma. Let $X$ be a metrizable space. If $\mathscr{A}$ is an open covering of $X$, then there is an open covering $\mathscr{E}$ of $X$ refining $\mathscr{A}$ that is countably locally discrete.

Lemma. Let $X$ be a regular space with a basis $\mathscr{B}$ that is countably locally discrete. Then $X$ is a $T_6$ space. That is to say, $X$ is normal, and every closed set in $X$ is a $G_\delta$ set in $X$.

Actually the proof of the original version of the last two lemmas in the textbook is already strong enough to prove the discrete version.

§41 Paracompactness

Ex.41.1

Give an example to show that if $X$ is paracompact, it does not follow that for every open covering $\mathscr{A}$ of $X$, there is a locally finite subcollection of $\mathscr{A}$ that covers $X$.

Endow $\Bbb Z$ with the order topology (the same as discrete topology), then $\lbrace(-n,n)\rbrace_n$ is an open covering of $\Bbb Z$ without locally finite subcovering

*Actually, we can show the following:

TFAE
(1) $X$ is compact.
(2) Every open covering of $X$ has a locally finite subcovering.
(3) Every open covering of $X$ has a point finite subcovering.

The direction $(1)\implies(2)\implies(3)$ is obvious. For $(3)\implies(1)$:

Let $\mathscr{A}=\lbrace U_\alpha\rbrace$ be an open covering of $X$. Pick any element $U\in\mathscr{A}$ and form a new collection $\mathscr{A}’=\lbrace U\cup U_\alpha|U_\alpha\in\mathscr{A}\rbrace$. It is clear that $\mathscr{A}’$ is an open covering of $X$. Then $\mathscr{A}’$ has a point finite subcovering $\mathscr{A}’’$. Now pick any point $x\in U$. We can see every element of $\mathscr{A}’’$ contains $x$. But $\mathscr{A}’’$ is point finite. Therefore $\mathscr{A}’’$ is finite, say $\mathscr{A}’’=\lbrace U\cup U_\alpha|\alpha\in F\rbrace$ where $F$ is finite. Then $U, U_\alpha,(\alpha\in F)$ covers $X$. Hence $X$ is compact.

Ex.41.2

(a) Show that the product of a paracompact space and a compact space is paracompact. [Hint: Use the tube lemma.]
(b) Conclude that $S_\Omega$ is not paracompact.

(a) Let $X,Y$ be paracompact and compact respectively, and $\mathscr{A}$ be an open covering of $X\times Y$. Since $Y$ is compact, for each $x\in X $, there exists finitely many open sets $U_n(x)\in\mathscr{A}$ that cover $\lbrace x\rbrace\times Y$. By the tube lemma, each $U_n(x)$ actually contains a tube $U_x\times Y$ where $U_x\in\mathcal N(x)$. Now, $\lbrace U_x\rbrace$ forms an open covering of $X$, thus having an open locally finite refinement $\mathscr{B}$ that covers $X$.

For each $V\in\mathscr{B}$, choose an $x=f(V)\in X$ such that $V\subset U_x$. Then, $\lbrace U_n(f(V))\cap (V\times Y)\rbrace_{V\in\mathscr{B},n}$ is an open covering of $X\times Y$ that is locally finite and refines $\mathscr{A}$, as you can readily check.

(b) Suppose on the contrary $S_\Omega$ is paracompact. Then $S_\Omega\times \overline{S_\Omega}$ is paracompact, by part (a). However, this will lead us to conclude that $S_\Omega\times \overline{S_\Omega}$ is $T_4$ (normal), which is false.

Ex.41.3

Is every locally compact Hausdorff space paracompact?

No, consider $S_\Omega$. We have just proven that it is not paracompact.

Ex.41.4

(a) Show that if $X$ has the discrete topology, then $X$ is paracompact.
(b) Show that if $f:X\to Y$ is continuous and $X$ is paracompact, the subspace $f(X)$ of $Y$ need not be paracompact.

(a) If $\mathscr{A}$ is an open covering of $X$, then $\lbrace \lbrace x\rbrace\subset X|\exists U\in\mathscr{A},x\in U\rbrace$ is an open refinement of $\mathscr{A}$ that covers $X$.

(b) By part (a), we just need to find a non-paracompact space $X$, then consider the identity map $\mathbb 1: X(\mathrm{Discrete})\to X$. For example, $S_\Omega$.

Ex.41.5

Let $X$ be paracompact. We proved a “shrinking lemma” for arbitrary indexed open coverings of $X$. Here is an “expansion lemma” for arbitrary locally finite indexed families in $X$.
Lemma. Let $\lbrace B_\alpha\rbrace_{\alpha\in J}$ be a locally finite indexed family of subsets of the paracompact Hausdorff space $X$. Then there is a locally finite indexed family $\lbrace U_\alpha\rbrace_{\alpha\in J}$ of open sets in $X$ such that $B_\alpha\subset U_\alpha$ for each $\alpha$.

We mimic the trick in Lemma 41.3, part $(3)\implies(4)$.

Write $\mathscr{B}=\set{B_\alpha}$. For each point $x\in X$, there is a neighborhood of $x$ that intersects only finitely many elements of $\mathscr{B}$. The collection of all open sets that intersect only finitely many elements of $\mathscr{B}$ is thus an open covering of $X$. There exists an locally finite open refinement of this collection that covers $X$, then form a collection $\mathscr{C}$ that consists of closures of elements in this refinement. Then $\mathscr{C}$ is an locally finite closed covering of $X$.

Next, for each $B_\alpha\in\mathscr{B}$, let $E(B_\alpha)=X\setminus\bigcup_{C\in\mathscr{C}\text{ and }C\subset X\setminus B_\alpha}C$. Now follow the same argument in Lemma 41.3, you know $\mathscr{U}=\set{U_\alpha}:=\set{E(B_\alpha)|\alpha\in J}$ is a locally finite open collection that covers $X$, satisfying $B_\alpha\subset U\alpha$ for each $\alpha$.

Ex.41.6

(a) Let $X$ be a regular space. If $X$ is a countable union of compact subspaces of $X$, then $X$ is paracompact.
(b) Show $\Bbb R^\infty$ is paracompact as a subspace of $\Bbb R^\omega$ in the box topology.

(a) i.e. show that in $T_3$ cases, $\sigma$-compact implies paracompact.

Let $X=\bigcup C_n$ where $C_n$ are compact in $X$. If $\mathscr{A}$ is an open covering of $X$, for each $n$, there are finitely many elements in $\mathscr{A}$ that covers $C_n$, denote it as $\mathscr{D}_n$. Obviously $\mathscr{D}_n$ is locally finite (since finite) so $\mathscr{D}=\bigcup\mathscr{D}_n$ is countably locally finite and is an open covering of $X$. Thus, every open covering has a countably locally finite open refinement that covers $X$. By Lemma 41.3, $X$ is paracompact.

Note that we actually proved $X$ is Lindelof!

(b) $\Bbb R^\infty=\bigcup_n \Bbb R^n_*$, where $\Bbb R^n_*=\set{(x_i)_i\in\Bbb R^\omega|x_i\in\Bbb R\text{ for }i\le n, x_i=0\text{ otherwise}}\cong\Bbb R^n$ is compact.

That $\Bbb R^\infty$ is $T_3$ (even $T_{3.5}$) in box topology is given in Ex.33.9, using the Urysohn Lemma.

Ex.41.7

Let X be a regular space.
(a) If $X$ is a finite union of closed paracompact subspaces of $X$, then $X$ is paracompact.
(b) If $X$ is a countable union of closed paracompact subspaces of $X$ whose interiors cover $X$, show $X$ is paracompact.

(a) Clearly we only need to show the case $n=2$. Suppose $X=A_1\cup A_2$ where $A_i$ is closed paracompact subspace of $X$. For any open covering $\mathscr{U}$ of $X$, the collection $\mathscr{U}_i=\set{U\cap A_i|U\in\mathscr{U}}$ is an open covering of $A_i$, thus have a locally finite closed refinement $\mathscr{U}_i’$. Since $A_i$ is closed, every element in $\mathscr{U}_i’$ is also closed in $X$. So $\mathscr{V}=\mathscr{U}_1’\cup\mathscr{U}_2’$ is a closed covering of $X$.

For each $x\in X$, if $x\in A_1$ then it has an open nbhd $U$ in $A_1$, such that $U$ intersects finitely many sets in $\mathscr{U}_1’$. There exists an open set $U’$ of $X$ such that $U=U’\cap A_1$. Similarly if $x\in A_2$ there exists an open set $V’$ of $X$ such that $V=V’\cap A_2$ intersects finitely many sets in $\mathscr{U}_2’$. Note that $U’$ still intersects only finitely many elements in $\mathscr{U}_1’$ and similar to $V’$

Now consider two open sets $U’\setminus A_2, V’\setminus A_1$ of $X$. Note that $U’\setminus A_2\subset U, V’\setminus A_1\subset V$, and neither $U’\setminus A_2$ intersect any element in $\mathscr{U}_2’$ nor $V’\setminus A_1$ intersect any element in $\mathscr{U}_1’$. So, if $x\notin A_1$ then $W=U’\setminus A_2$ is a open nbhd of $x$ that intersects only finitely many elements in $\mathscr{V}$, and $x\notin A_1$ is similar. In the case $x\in A_1\cap A_2$, choose $W=U’\cap V’$ and note that $W\subset U’,V’$. Thus $\mathscr{V}$ is locally finite. By Lemma 41.3, $X$ is paracompact.

*This problem is not hard, but also not so obvious like some readers might think. Note that unlike compactness, it is NOT correct that a subspace $Y\subset X$ is paracompact iff every open covering of $Y$ consisting of open sets of $X$ has locally finite open (in the sense of $X$) refinement that covers $Y$.

(b) Let $\mathscr{U}$ be an open covering of $X=\bigcup X_n$ with $\bigcup(X_n)^\circ=X$.

For each $n$,
$\mathscr{U}_n=\set{U\cap X_n|U\in\mathscr{U}}$ is an open covering of $X_n$.

Thus there is a family $\mathscr{V}_n’$ of open subsets of $X$ which has the property that $\set{V\cap X_n|V\in\mathscr{V}’_n}$ is a locally-finite open refinement (in $X_n$) of $\mathscr{U}_n$. Write $\mathscr{V}_n=\set{V\cap(X_n)^\circ|V\in\mathscr{V}’_n}$ to obtain a locally-finite family of open subsets of $X$ which covers the interior $(X_n)^\circ$ and refines $\mathscr{U}_n$. The collection $\mathscr{V}=\bigcup\mathscr{V}_n$ is now a countably locally finite open refinement of $\mathscr{U}$ covering $X$. By Lemma 41.3, $X$ is paracompact

Ex.41.8

Let $p:X\to Y$ be a perfect map. (See Exercise 7 of §31.)
(a) Show that if $Y$ is paracompact, so is $X$. [Hint: If $\mathscr{A}$ is an open covering of $X$, find a locally finite open covering of $Y$ by sets $B$ such that $p^{-1}(B)$ can be covered by finitely many elements of $\mathscr{A}$; then intersect $p^{-1}(B)$ with these elements of $\mathscr{A}$.]
(b) Show that if $X$ is a paracompact Hausdorff space, then so is $Y$. [Hint: If $\mathscr{B}$ is a locally finite closed covering of $X$, then $\set{p(B)|B\in\mathscr{B}}$ is a locally finite closed covering of $Y$.]

(a) Recall that perfect map is a surjective continuous map that is closed with compact fibers. If $\mathscr{A}$ is an open covering of $X$. WLOG we may assume $\mathscr{A}$ is closed under finite unions. Let $\mathscr{B}=\set{Y\setminus f(X\setminus A)|A\in\mathscr{A}}$. It is clear that $\mathscr{B}$ is a collection of open sets of $Y$. And since $f$ has compact fibers and $\mathscr{A}$ is closed under finite unions, each fiber $f^{-1}(y)$ is contained in an $A\in\mathscr{A}$. Thus,
$$
\begin{aligned}
X\setminus f^{-1}(y)&\supset X\setminus A\\
f(X\setminus f^{-1}(y))&\supset f(X\setminus A)\\
Y\setminus f(X\setminus f^{-1}(y))&\subset Y\setminus f(X\setminus A)\\
=Y\setminus f f^{-1}(Y\setminus\set{y})&\\
=\set{y}&
\end{aligned}
$$
Therefore, $\mathscr{B}$ covers $Y$. So it has a locally finite open refinement $\mathscr{B}’$ that covers $Y$. Now let $\mathscr{A}’=\set{f^{-1}(B’)|B’\in\mathscr{B}’}$. This is an open covering of $Y$. It refines $\mathscr{A}$, because for each $A’=f^{-1}(B’)\in\mathscr{A}’$ where $B’\in\mathscr{B}’$, there exists an $A\in\mathscr{A}$ such that $B’\subset Y\setminus f(X\setminus A)$. Hence,
$$
\begin{aligned}
A’&\subset f^{-1}(Y\setminus f(X\setminus A))\\
&= X\setminus f^{-1}f(X\setminus A)\\
&\subset A
\end{aligned}
$$
Therefore, $\mathscr{A}’$ refines $\mathscr{A}$. As for locally finiteness, for each $x\in X,\exists V\ni f(x)$, s.t. $V$ intersects finitely many $B’\in\mathscr{B}’$, then $f^{-1}V$ intersects those $f^{-1}B’$. Moreover, if $f^{-1}V\cap f^{-1}B’\neq\emptyset$, we have $\emptyset\neq f(f^{-1}V\cap f^{-1}B’)\subset V\cap B’$, meaning that $f^{-1}V$ will only intersect those elements, thus locally finite.

(b) Recall in Ex.31.7 we have proved that if $f:X\to Y$ is perfect map, then $X$ regular implies $Y$ regular. Here $X$ is paracompact Hausdorff, hence regular, hence $Y$ is regular. Let $\mathscr{U}$ be an open covering of $Y $, take the preimages to form an $\mathscr{A}$, which has a closed locally finite refinement $\mathscr{A}’$ that covers $X$. Let $\mathscr{B}$ be the images of elements of $\mathscr{A}’$.

Now clearly $\mathscr{B}$ is a closed covering of $Y$. It refines $\mathscr{A}$, for if $B\in\mathscr{B}$, there exists $A’\in\mathscr{A}’,A\in\mathscr{A},U\in\mathscr{U}$ such that: $B=f(A’)\subset f(A)=ff^{-1}(U)=U$. Now we prove locally finiteness.

For each $y\in Y$, its fiber is compact, thus can be covered by finitely many open sets, each of which intersects finitely many elements in $\mathscr{A}’$, i.e. $f^{-1}y\subset M=\bigcup A_i’$ . Therefore, $y\in f(\bigcup A_i’)$. Closed maps send nbhd to nbhd (in particular, in part (a) we proved $N=Y\setminus f(X\setminus A)\subset f(A)$), so there exists $N$ such that $y\in N\subset f(M)$. Similarly we can prove $f^{-1}N\subset M$. Hence $f^{-1}N$ intersects finitely many elements of $\mathscr A’$. Similar to the end of part (a), we know $N$ intersects at most finitely many elements of $\mathscr B$.

Ex.41.10

Theorem. If $X$ is a Hausdorff space that is locally compact and paracompact, then each component of $X$ has a countable basis.
Proof. If $X_0$ is a component of $X$, then $X_0$ is locally compact and paracompact. Let $\mathscr{C}$ be a locally finite covering of $X_0$ by sets open in $X_0$ that have compact closures. Let $U_1$ be a nonempty element of $\mathscr{C}$, and in general let $U_n$ be the union of all elements of $\mathscr{C}$ that intersect $\overline{U_{n-1}}$. Show $\overline{U_n}$ is compact, and the sets $U_n$ cover $X_0$.

THIS IS FALSE. A correction may be changing ‘paracompact’ to ‘metrizable’.

After correction, it is equivalent to proving locally compact connected metrizable space is $C_2$. We show $X$ is a countable union of $C_2$ open subspaces, thus is $C_2$.

We follow the hint. Since $X$ is locally compact, those open sets with compact closures form an open covering. Let $\mathscr{C}$ be a locally finite open refinement of it covering $X$ and $\emptyset\neq U_1\in\mathscr{C}$, $U_2=\bigcup\set{U\in\mathscr{C}|U\cap\overline{U_1}\neq\emptyset}$,…,$U_{n+1}=\bigcup\set{U\in\mathscr{C}|U\cap\overline{U_n}\neq\emptyset}$. We prove by induction that $\overline{U_n}$ is compact. It is clear that $\overline{U_1}$ is compact. Now suppose $\overline{U_n}$ is compact. Since $\mathscr{C}$ is locally finite, for each $x\in\overline{U_n}$, there is an open nbhd $U_x$ of $x$ such that $U_x$ intersects only finitely many elements in $\mathscr{C}$. Since $\overline{U_n}$ is compact, we can cover it by finitely many such $U_x$. In particular, $\overline{U_n}$ intersects only finitely many elements of $\mathscr{C}$. By definition, we know that $U_{n+1}$ is a finite union of $U\in\mathscr{C}$. Therefore, $\overline{U_{n+1}}$ is compact. Hence we conclude all $\overline{U_n}$ are compact.

Now we show $X=\bigcup U_n$, where connectedness is used. Let $x\in\overline{\bigcup U_n}$. Every open nbhd of it intersects $\bigcup U_n$, or equivalently, intersects some $U_n$, hence $\overline{U_n}$. By definition, $x\in U_{n+1}$. Therefore, $x\in\bigcup U_n$, i.e. $\overline{\bigcup U_n}=\bigcup U_n$. Hence $\bigcup U_n$ is a non-empty clopen set. By connectedness, $\bigcup U_n=X$

Finally, we have $X=\bigcup\overline{U_n}$. Each $\overline{U_n}$ is a compact metrizable space, hence $C_2$. Being open subspace of $\overline{U_n}$, $U_n$ is $C_2$. Thus we come to our conclusion.

§42 The Smirnov Metrization Theorem

Ex.42.1

Compare Theorem 42.1 with Exercises 7 and 8 of §34.

There are three theorems concerning local metrizability. I list them below:

In the following spaces, local metrizability is equivalent to metrizability.
(1) compact Hausdorff spaces (Ex.34.7)
(2) Lindelof regular spaces (Ex.34.8)
(3) paracompact Hausdorff spaces (Theorem 42.1)

Ex.42.2

(a) Show that for each $x\in S_\Omega$, the section of $S_\Omega$ by $x$ has a countable basis and hence is metrizable.
(b) Conclude that $S_\Omega$ is not paracompact.

(a) For each $x\in S_\Omega$, denote $S_x$ its section. All open intervals in $S_x$ form a basis of $S_x$. The cardinality is $\aleph_0^2=\aleph_0$, hence $S_x$ is $C_2$. By the Urysohn lemma, $S_x$ is metrizable.

(b) By part (a), $S_\Omega$ is locally metrizable. If moreover $S_\Omega$ is paracompact, then it is metrizable. However, this is false. (Recall that $S_\Omega$ is not compact but is limit point compact).