Hausdorff paracompact space is normal

Question

Studied from Munkres' Topology its proof. He says that the Hausdorff condition enables us to choose, for each $b\in B$ an open set $U_b$ whose closure is disjoint from $a$. I did not get why disjoint closure is needed is it not enough to show that $U_b$ is disjoint from $a$? And similar question at the end when he says that closure of $V$ is disjoint from $a$.

Answer 1

Let's review the Munkres' proof: the assumptions are that $X$ is paracompact and $T_2$ (Hausdorff).

Step 1: show $X$ is regular (and thus $T_3$), so let $B$ be closed and $a \notin B$.

Then for each $b \in B$ we pick an open (by Hausdorffness!) disjoint set $U_b$, that contains $b$ and an open $O$ that contains $a$. As $O \cap U_b = \emptyset$, we see that $a \notin \overline{U_b}$ ($a$ is in the closure of a set if every neighbourhood of $a$ intersects that set, and here we have a witness $O$ that shows that that is not the case for $a$ and $U_b$). We forget about these $O$ and we only keep all the sets $U_b$ with $a \notin \overline{U_b}$, one for each $b \in B$.

This defines an open cover $\mathcal{U}=\{U_b: b \in B\} \cup \{X- B\}$ of $X$ (here we use that $B$ is closed; this is clearly a cover of $X$ as all $b \in B$ are covered by their "own" $U_b$ and the rest by $X-B$.

So by paracompactness there is an open refinement $\mathcal{C}$ that is locally finite, and then Munkres defines $$\mathcal{D}=\{C \in \mathcal{C}: C \cap B \neq \emptyset\}$$

and this collection of open sets is also locally finite (as a subset of $\mathcal{C}$, see 39.1) and any $D \in \mathcal{D}$ is a subset of some $U_b$ (from being a refinement of $\mathcal{U}$ and intersecting $B$) and so $a \notin \overline{D} \subseteq \overline{U_b}$ as well. And so $$V= \bigcup \{D \in \mathcal{D}\}$$ is open, contains $B$, and by 39.1 we also have that

$$\overline{V}=\bigcup\{\overline{D}: D \in \mathcal{D}\}$$

does not contain $a$. So $a \in X-\overline{V}$ and $B \subseteq V$ are open disjoint neighbourhoods of $a$ and $B$ respectively, proving regularity.

Step 2. $X$ is normal:

let $A$ and $B$ be closed and disjoint. Then by step 1 we can find for each $b \in B$ open disjoint subsets $O$ and $U_b$ such that

$$b \in U_b, A \subseteq O$$ and this implies that $$A \cap \overline{U_b}=\emptyset$$

as every point in $A$ has a neighbourhood $O$ that misses $U_b$, etc.

Again we define $\mathcal{U}$ as above, its locally finite refinement $\mathcal{C}$ and subcollection $\mathcal{D}$ of members that intersect $B$.

$V$ as defined above as union of $\mathcal{D}$ is then an open neighbourhood of $B$ whose closure misses $A$ (instead of $a$ as before), and we finish the same way too to get normality. Instead of $a \notin \overline{U_b}$ we use $A \cap \overline{U_b}=\emptyset$, whenever relevant.

It's basically the same subproof twice: first to get from point-point separation to point-closed set and finally from that to closed-closed. The proof that compact Hausdorff spaces are normal (of which this fact is a generalisation) goes along the same route. The finite subcover can be replaced by localy finite using the closure-preserving property of locally finite collections (39.1, point 2).

Answer 2

It may not be needed in some particular contexts, but it is exactly what the Hausdorff $T_2$ condition gives you: if you imagine an open neighbourhood $U_b$ disjoint from $a$ but whose closure contains $a$, then it is impossible to find a neighbourhood of $a$, $U_a$, such that $U_a\cap U_b=\emptyset$.

The $T_2$ condition ensures that there should be such a $U_a$ for some $U_b$.

Answer 3

One point sets are closed in Hausdorf space. Can you apply your logic to one point sets? You will find the answer.