5.20 Locally Compact Hausdorff Spaces
Definition 5.20.1 (Locally Compact Hausdorff Space).label Let $X$ be a Hausdorff space, then the following are equivalent:
- (1)
For any $x \in X$, there exists $K \in \cn(x)$ compact.
- (2)
For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of compact sets.
- (3)
For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of precompact sets.
If the above holds, then $X$ is a locally compact Hausdorff (LCH) space.
Proof. (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By Proposition 5.16.3, $K$ itself is closed, so $\overline{U \cap K}\subset K$ is a closed subset of a compact set, and compact by Proposition 5.16.2.
(2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By Proposition 5.16.3, $K$ is closed, so $\overline{K^o}\subset K$ is compact by Proposition 5.16.2.$\square$
Lemma 5.20.2.label Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^{o}(K)$ precompact such that $K \subset V \subset \ol{V}\subset U$.
Proof. For each $x \in K$, there exists $V_{x} \in \cn^{o}(x)$ be precompact such that $x \in V_{x} \subset \overline{V_x}\subset U$ by (3) of Definition 5.20.1. Since $K$ is compact, there exists $\seqf{x_j}\subset K$ such that
so $V = \bigcup_{j = 1}^{n} V_{x_j}\in \cn^{o}(K)$ is precompact.$\square$
Lemma 5.20.3 (Urysohn’s Lemma (LCH)).label Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_{c}(X; [0, 1])$ such that $\supp{f}\subset U$.
Proof [Lemma 4.32, Fol99]. By Lemma 5.20.2, there exists $V, W \in \cn^{o}(K)$ precompact such that
As $\ol{W}$ is compact, it is normal by Proposition 5.16.5. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by Proposition 5.16.3.
By Urysohn’s lemma, there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_{K} = 1$ and $f|_{\ol{W} \setminus V}= 0$. Let
then by the gluing lemma for continuous functions, $F \in C_{c}(X; [0, 1])$ with $F|_{K}= 1$ and $\supp{f}\subset U$.$\square$
Theorem 5.20.4 (Tietze Extension Theorem (LCH)).label Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^{o}(K)$, and $f \in C(K; \real)$, then there exists $F \in C_{c}(U; \real)$ such that $F|_{K} = f$.
Proof. By Lemma 5.20.2, there exists $V, W \in \cn^{o}(K)$ precompact such that $K \subset V \subset \ol{V}\subset U$. As $\ol{W}$ is compact, it is normal by Proposition 5.16.5. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by Proposition 5.16.3.
By the Tietze extension theorem, there exists $F \in C(\ol{W}; \real)$ such that $F|_{K} = f$. By Urysohn’s lemma, there exists $\eta \in C_{c}(X; [0, 1])$ such that $\eta|_{K} = 1$ and $\supp{\eta}\subset V$. In which case, define
then by the gluing lemma for continuous functions, $\ol F \in C_{c}(X; \real)$ with $\ol F|_{K} = F|_{K} = f$ and $\supp{F}\subset \supp{\eta}\subset V \subset U$.$\square$
Proposition 5.20.5.label Let $X$ be a LCH space, then:
- (1)
$X$ is compactly generated.
- (2)
For any uniform space $Y$, $C(X; Y) \subset Y^{X}$ is closed with respect to the compact-open topology.
Proof. (1): Let $U \subset X$ such that $U \cap K$ is open in $K$ for all $K \subset X$ compact. For any $x \in U$, there exists a compact neighbourhood $K \in \cn(x)$. In which case, $U \supset U \cap K \in \cn(x)$, so $U \in \cn(x)$ for all $x \in U$. By Lemma 5.4.3, $U$ is open.
(2): By Proposition 7.2.6, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$. Since $X$ is compactly generated, $C(X; Y) \subset Y^{X}$ is closed with respect to the compact-open topology by Corollary 7.3.3.$\square$
Proposition 5.20.6.label Let $\seqi{X}$ be a family of LCH spaces. If $X_{i}$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_{i}$ is a LCH space.
Proof. By Proposition 5.8.3, $\prod_{i \in I}X_{i}$ is Hausdorff. Let $x \in \prod_{i \in I}X_{i}$ and $i \in I$. If $X_{i}$ is not compact, let $U_{i} \in \cn_{X_i}(\pi_{i}(x))$ be compact. Otherwise, let $U_{i} = X_{i}$. Let $U = \prod_{i \in I}U_{i}$, then since $U_{i} \ne X_{i}$ for only finitely many $i \in I$, $U \in \cn_{X}(x)$. By Tychonoff’s Theorem, $U$ is compact. Therefore $X$ is locally compact.$\square$
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