4.3Normal Spaces

Proof.

Suppose $X$ is a regular, second-countable space. Then $X$ has a countable basis $\mathcal{B}$. Let $A, B$ be any two closed subsets of $X$. Since $A$ is closed, for any any $x \in A$, we may choose an open neighborhood $U$ of $x$ that is disjoint from $B$. By regularity, we know there is a neighborhood $V$ of $x$ for which $\bar{V} \subseteq U$. Since $V$ is open, there is a basis element $O \in \mathcal{B}$ such that $x \in O \subseteq V$. Obviously, $\bar{O} \cap B = \emptyset$, since $\bar{O} \subseteq \bar{V}$. Identify each point $x$ with some basis element as such, then let $\{U_n\}$ be the collection of all of these basis elements (note this is a subset of the countable basis, and so it is countable). Applying the same procedure to $B$, let the collection be $\{V_n\}$.

Now note $\bigcup U_n, \bigcup V_n$ are open sets containing $A, B$, but they are not necessarily disjoint. Instead, let us construct two open covers that are disjoint. For each $n$, define

Note the finite union of compact spaces is compact, and as subspaces of a Hausdorff space, it is closed. Note each $U_n', V_n'$ is a complement of a closed set with respect to the open sets $U_n, V_n$, so they are open. Moreover, for any $x \in A$, there is some $n \in \bZ_+$ where $x \in U_n$ and $x \notin \bar{V_i}$ for each $i \in \{1, \dots, n\}$ (obviously, by $\bar{V_i}$ being disjoint from $A$), and vice-versa holds for $B$.

We have shown $U' = \bigcup U_n'$ and $V' = \bigcup V_n'$ are open sets containing $A$ and $B$, and now we show they are disjoint. For the sake of contradiction, suppose $x \in U' \cap V'$, and so $x \in U_j' \cap V_k'$ for some $j, k \in \bZ_+$. If $j \le k$, then by definition of $U_j'$, we have that $x \in U_j$; by definition of $V_k'$, we have that $x \notin \bar{U_j}$, a contradiction. ❦

Proof.

Let $(X, d)$ be a metric space, and let $A, B \subseteq X$ be disjoint, closed sets. For each $a \in A$, there is some $\epsilon_a > 0$ such that $B(a, \epsilon_a) \cap B = \emptyset$, since otherwise $a$ would be a limit point of $B$, forcing $a \in B$ by $B$ being closed. An analogous procedure can be used to find $\epsilon_b$ for each $b \in B$. Finally, define

with $\epsilon_a, \epsilon_b$ as chosen before. Obviously, $U$ and $V$ form open sets that contain $A$ and $B$. We prove they are disjoint by contradiction. Suppose $z \in U \cap V$, and so there exist $a \in A$ and $b \in B$ such that $z \in B(a, \epsilon_a/2) \cap B(b, \epsilon_b/2)$. Note that

If $\epsilon_a \le \epsilon_b$, then $d(a, b) < \epsilon_b$, and so $a \in B(b, \epsilon_b/2)$. Vice-versa is true if $\epsilon_b \ge \epsilon_b$. But note the balls about points in $B$ are disjoint from $A$ by construction, and vice-versa for the balls constructed about points in $A$. Thus we have reached a contradiction, and so $U \cap V = \emptyset$, completing the proof. ❦

Proof.

Suppose $X$ is compact Hausdorff. We first show $X$ is regular. Let $x$ be any point in $X$ and $B$ be any disjoint, closed set. For any $b \in B$, we have the existence of open neighborhoods $U_b$ and $V_b$ of $x$ and $b$ that are disjoint. Note that $\{V_b\}$ is an open cover for $B$ from open sets of $X$, and since $B$ is a closed subset of a compact space, there is a finite subcover $\{V_k\}_{k = 1}^n$. Then $U = \bigcup_{b \in B} U_b$ and $V = \bigcap_{k = 1}^n V_k$ are disjoint, open sets containing $x$ and $B$ respectively. (Alternatively, note $X$ is compact Hausdorff and thus locally compact Hausdorff, then apply a previous theorem.)

For any two disjoint, closed sets $A, B$, produce disjoint, open neighborhoods for each $a \in A$ and $B$. Intersect a finite subcover of the open neighborhoods covering $B$, then note it is an open neighborhood disjoint from the open neighborhood about $A$, which is the union of each open neigbhorhood about every $a \in A$. ❦