5.9 Regular Spaces
Definition 5.9.1 (Regular Space).label Let $X$ be a topological space, then the following are equivalent:
- (1)
For each $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U \in \cn(x)$ and $V \in \cn(A)$ such that $U \cap V = \emptyset$.
- (2)
For each $x \in X$, the closed neighbourhoods of $x$ forms a fundamental system of neighbourhoods at $x$.
If $X$ is a T1 space such that the above holds, then $X$ is regular.
Proof [Proposition 1.4.11, Bou13]. $(1) \Rightarrow (2)$: Let $U \in \cn^{o}(x)$, then $U^{c}$ is closed with $x \not\in U^{c}$ by (V1). Thus there exists $V \in \cn^{o}(U^{c})$ such that $x \not\in V$, so $V^{c} \in \cn(x)$ is closed with $V^{c} \subset U$.
$(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^{c} \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^{c}$. Thus $K \in \cn(x)$ and $K^{c} \in \cn(A)$ are the desired neighbourhoods.$\square$
Theorem 5.9.2 ([Theorem 1.8.1, Bou13]).label Let $X$ be a topological space, $Y$ be a regular space, $A \subset X$, and $f \in C(A; Y)$, then the following are equivalent:
- (1)
There exists $F \in C(X; Y)$ such that $F|_{A} = f$.
- (2)
For each $x \in X$, $\lim_{y \to x, y \in A}f(y)$ exists.
Proof. $(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_{A} = f$ by (4) of Definition 5.8.1.
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of Definition 5.9.1, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^{o}(x)$ such that $f(U \cap A) \subset V$. In which case, Lemma 5.4.3 implies that $U \in \cn^{o}(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.$\square$
Proposition 5.9.3.label Let $X$ be a second countable regular space, then $X$ is normal.
Proof. Let $\cb \subset 2^{X}$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_{x}, V_{y} \in \cb$ such that $x \in \ol{U_x}\subset B^{c}$ and $y \in \ol{V_y}\subset B^{c}$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively.
For each $n \in \natp$, let
then $U_{n}'$ and $V_{n}'$ are both open. Let
then $U \in \cn_{X}(A)$ and $V \in \cn_{X}(B)$. For each $m, n \in \natp$ with $m \le n$, $V_{n} \cap \bigcup_{j = 1}^{n} U_{j} = \emptyset$, so $U_{m} \cap V_{n} = \emptyset$. Likewise, if $m \ge n$, then $U_{m} \cap V_{n} = \emptyset$ as well. Therefore $U \cap V = \emptyset$.$\square$
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