[Munkres Topology] 32. Normal Spaces

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Theorem 32.1. Every regular space with a countable basis is normal.

Proof. Let $X$ be a regular space with a countable basis $\mathcal{B}$. Let $A$ and $B$ be disjoint closed subsets of $X$. Each point $x$ of $A$ has a neighborhood $U$ not intersecting $B$. Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U$ (intersect with $U$.); finally, choose an element of $\mathcal{B}$ containing $x$ and contained in $V$. By choosing such a basis element for each $x$ in $A$, we construct a countable covering of $A$ by open sets whose closures do not intersect $B$(??? since $B$ is closed). Since this covering of $A$ is countable, we can index it with the positive integers; let us denote it by $\{U_n\}$.

Similarly, choose a countable collection $\{V_n\}$ of ope sets covering $B$, such that each set $\bar{V}_n$ is disjoint from $A$. The sets $U=\bigcup U_n$ and $V=\bigcup V_n$ are open sets containing $A$ and $B$, respectively, but they need not be disjoint. We perform the following simple trick to construct two open sets that are disjoint. Given $n$, define $$U'_n=U_n-\bigcup_{i=1}^n \bar{V}_i\quad \mbox{and}\quad V'_n=V_n -\bigcup_{i=1}^n \bar{U}_i.$$ Note that each set $U'_n$ is open, being the difference of an open set $U_n$ and a closed set $\bigcup_{i=1}^n \bar{V}_i$. Similarly, each set $V'_n$ is open. The collection $\{U'_n\}$ covers $A$, because each $x$ in $A$ belongs to $U_n$ for some $n$, and $x$ belongs to none of the sets $\bar{V}_i$. Similarly, the collection $\{V'_n\}$ covers $B$. See Figure 32.1.

Finally, the open sets $$U'=\bigcup_{n\in \mathbb{Z}_+} U'_n\quad \mbox{and}\quad V'=\bigcup_{n\in \mathbb{Z}_+} V'_n$$ are disjoint. For if $x\in U'\cap V'$, then $x\in U'_j \cap V'_k$ for some $j$ and $k$. Suppose that $j\le k$. It follows from the definition of $U'_j$ that $x\in U_j$; and since $j\le k$ it follows from the definition of $V'_k$ that $x\notin \bar{U}_j$. A similar contradiction arises if $j\ge k$. wow..

Theorem 32.2. Every metrizable space is normal.

Proof. Let $X$ be a metrizable space with metric $d$. Let $A$ and $B$ be disjoint closed subsets of $X$. For each $a\in A$, choose $\epsilon_a$ so that the ball $B(a,\epsilon_a)$ does not intersect $B$. Similarly, for each $b$ in $B$, choose $\epsilon_b$ so that the ball $B(b,\epsilon_b)$ does not intersect $A$. Define $$U=\bigcup_{a\in A} B(a, \epsilon_a/2)$\quad \mbox{and}\quad V=\bigcup_{b\in B} B(b,\epsilon_b/2).$$ Then $U$ and $V$ are open sets containing $A$ and $B$, respectively; we assert they are disjoint. For if $z\in U\cap V$, then $$z \in B(a, \epsilon_a/2)\cap B(b, \epsilon_b/2)$$ for some $a\in A$ and some $b\in B$. The triangle inequality applies to show that $d(a,b)<(\epsilon_a + \epsilon_b)2$. If $\epsilon_a \le \epsilon_b$, then $d(a,b)<\epsilon_b$, so that the ball $B(b,\epsilon_b)$ contains the point $a$. If $\epsilon_b \le \epsilon_a$, then $d(a,b)<\epsilon_a$, so that the ball $B(a,\epsilon_a)$ contains the point $b$. Neither situation is possible.

Theorem 32.3. Every compact Hausdorff space is normal.

Proof. Let $X$ be a compact Hausdorff space. We have already essentially proved that $X$ is regular. For if $x$ is a point of $X$ and $B$ is a closed set in $X$ not containing $x$, then $B$ is compact, so that Lemma 26.4 applies to show there exist disjoint open sets about $x$ and $B$, respectively.

Essentially the same argument as given in that lemma can be used to show that $X$ is normal: Given disjoint closed sets $A$ and $B$ in $X$, choose, for each point $a$ of $A$, disjoint open sets $U_a$ and $V_a$ containing $a$ and $B$, respectively. (Here we use regularity of $X$.) The collection $\{U_a\}$ covers $A$; because $A$ is compact, $A$ may be covered by finitely many sets $$U_{a_1},\cdots, U_{a_m}\quad \mbox{and} \quad V=V_{a_1}\cap \cdots \cap V_{a_m}$$ are disjoint open sets containing $A$ and $B$, respectively. (similar as Theorem 26.3.)

Theorem 32.4. Every well-ordered set $X$ is normal in the order topology.

Proof. Let $X$ be a well-ordered set. We assert that every interval of the form $(x,y]$ is open in $X$: If $X$ has a largest element and $y$ is that element, $(x,y]$ is just a basis element abour $y$. If $y$ is not the largest element of $X$, then $(x,y]$ equals the open set $(x,y')$, where $y'$ is the immediate successor of $y$. (Is it right for $\mathbb{R}$? um..)

Now let $A$ and $B$ be disjoint closed sets in $X$; assume for the moment that neither $A$ nor $B$ contains the smallest element $a_0$ of $X$. For each $a\in A$, there exists a basis element about $a$ disjoint from $B$; it contains some interval of the form $(x,a]$. (Here is where we use the fact that $a$ is not the smallest element of $X$.) Choose, for each $a\in A$, such an interval $(x_a,a]$ disjoint from $B$. Similarly, for each $b\in B$, choose an interval $(y_b,b]$ disjoint from $A$. The sets $$U=\bigcup_{a\in A} (x_a,a]\quad \mbox{and}\quad V=\bigcup_{b\in B} (y_b,b]$$ are open sets containing $A$ and $B$, respectively; we assert they are disjoint. For suppose that $z\in U\cap V$. Then $z\in (x_a,a]\cap (y_b,b]$ for some $a\in A$ and some $b\in B$. Assume that $a<b$. Then if $a\le y_b$, the two intervals are disjoint, while if $a>y_b$, we have $a\in (y_b,b]$, contrary to the fact that $(y_b,b]$ is disjoint from $A$. A similar contradiction occurs if $b<a$.

Finally, assume that $A$ and $B$ are disjoint closed sets in $X$, and $A$ contains the smallest element $a_0$ of $X$. The set $\{a_0\}$ is both open and closed in $X$.By the result of the preceding paragraph, there exist disjoint open sets $U$ and $V$ containing the closed sets $A-\{a_0\}$ and $B$, respectively. Then $U\cap \{a_0\}$ and $V$ are disjoint open sets containing $A$ and $B$, respectively.