4.20 Locally Compact Hausdorff Spaces

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 4.16.3, $K$ itself is closed, so $\overline{U \cap K}\subset K$ is a closed subset of a compact set, and compact by Proposition 4.16.2.

(2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By Proposition 4.16.3, $K$ is closed, so $\overline{K^o}\subset K$ is compact by Proposition 4.16.2.$\square$

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 4.20.1. Since $K$ is compact, there exists $\seqf{x_j}\subset K$ such that

By Proposition 4.5.4,

so $V = \bigcup_{j = 1}^{n} V_{x_j}\in \cn^{o}(K)$ is precompact.$\square$

Proof. By Lemma 4.20.2, there exists $V, W \in \cn^{o}(K)$ precompact such that

As $\ol{W}$ is compact, it is normal by Proposition 4.16.4. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by Proposition 4.16.3.

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 $\supp{f}\subset U$.$\square$

Proof. By Lemma 4.20.2, 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 4.16.4. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by Proposition 4.16.3.

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 $\supp{F}\subset \supp{\eta}\subset V \subset U$.$\square$

Proof. (1) $\Rightarrow$ (2): Let $\seq{K_n}\subset 2^{X}$ be compact such that $\bigcup_{n \in \natp}K_{n} = X$, and $U_{0} = \emptyset$.

Assume inductively that $\bracs{U_j}_{0}^{n}$ has been constructed such that:

1. For each $0 \le k \le n$, $U_{k}$ is a precompact open set.

2. For each $0 \le k < n$, $\overline{U_k}\subset U_{k+1}$.

3. For each $1 \le k \le n$, $U_{k} \supset \bigcup_{j = 1}^{k} K_{j}$.

By Lemma 4.20.2, there exists $U_{n+1}\in \cn^{o}(\overline{U_n}\cup K_{n+1})$ precompact. In which case, by (c),

Thus $\bracs{U_j}_{0}^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaustion of $X$ by compact sets.$\square$

Proof. Since $K$ is compact, assume without loss of generality that $\seqi{U}= \seqf{U_j}$.

For every $x \in K$, there exists $1 \le j \le n$ and $N_{x} \in \cn(x)$ compact such that $x \in N_{x} \subset U_{j}$. By compactness of $K$, there exists $\seqf[m]{x_j}\subset K$ such that $K = \bigcup_{j = 1}^{m} N_{x_j}$.

For each $1 \le j \le n$, let

then $F_{j} \subset U_{j}$ is compact, and $\bigcup_{j = 1}^{n} F_{j} \supset K$.

By Urysohn’s lemma, there exists $\seqf{f_j}\subset C_{c}(X; [0, 1])$ such that for each $1 \le j \le n$, $f_{j}|_{F_j}= 1$, and $\supp{f_j}\subset U_{j}$.

By Urysohn’s lemma again, there exists $f_{j + 1}\in C(X; [0, 1])$ such that $f_{j+1}|_{K}= 0$ and $\bracs{f_{j+1} = 0}\subset \bigcup_{j = 1}^{n} \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_{j}$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_{j} = f_{j}/F$, then $g_{j} \in C_{c}(X; [0, 1])$ with $\supp{g_j}\subset U_{j}$. In addition, since $f_{j+1}|_{K} = 0$,

Therefore $\seqf{g_j}$ is the desired partition of unity.$\square$

Proof. $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by Lemma 4.18.2. Let

then $F_{E} \in \cn(\ol{E})$ is precompact.

Let $N \subset X$ and $E \in \ce$. If $N \cap F_{E} \ne \emptyset$, then there exists $F \in \ce$ such that $N \cap F \ne \emptyset$ and $F \cap \ol{E}\ne \emptyset$. Thus

By Lemma 4.18.3, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.

Let $x \in X$, then there exists $N \in \cn(x)$ such that $\bracs{F \in \ce|N \cap F \ne \emptyset}$ is finite. In which case, $\bracs{E \in \ce|N \cap F_E \ne \emptyset}$ is finite as well. Therefore $\bracs{F_E}_{E \in \ce}$ is locally finite.

$(\bracs{G_E}_{E \in \ce})$: For each $x \in X$, there exists $E \in \ce$ and $N_{x} \in \cn^{o}(x)$ precompact with $x \in N_{x} \subset \ol{N_x}\subset E$.

For any $E \in \ce$, $\ol{E}$ is compact, so there exists $X_{E} \subset X$ finite such that

1. $\ol{E}\subset \bigcup_{x \in X_E}N_{x}$.

2. For every $x \in X_{E}$, $N_{x} \cap E \ne \emptyset$.

Let $X_{\ce}= \bigcup_{E \in \ce}X_{\ce}$, and for each $E \in \ce$, let

then $\bracs{G_E}_{E \in \ce}$ is an open cover of $X$. Since $G_{E} \subset E$ for all $E \in \ce$, $\bracs{G_E}_{E \in \ce}$ is locally finite.

It remains to show that $\ol{G_E}\subset E$. Let $x \in X_{F}$ such that $N_{x} \subset E$, then $N_{x} \cap F \ne \emptyset$. Since $N_{x} \subset E$, $E \cap F \ne \emptyset$. Thus

is finite by Lemma 4.18.2, so

by Proposition 4.5.4.$\square$

Proof. (1) $\Rightarrow$ (2): For each $x \in X$, there exists a precompact open neighbourhood $U_{x} \in \cn^{o}(x)$. Since $\bracs{U_x| x \in X}$ is an open cover of $X$, there exists a locally finite refinement $\mathcal{V}$. For each $V \in \mathcal{V}$, there exists $x \in X$ such that $V \subset U_{x}$. In which case, $\ol{V}\subset \ol{U_x}$ is compact.

(2) $\Rightarrow$ (3): Let $\cf \subset 2^{X}$ be a locally finite open cover of $X$ consisting of precompact open sets. By Lemma 4.20.7, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F}\subset G_{F}$ for all $F \in \cf$.

For each $F \in \cf$, let

then $\mathcal{U}_{F}$ is a precompact open cover of $\ol{F}$. By compactness of $\ol{F}$, there exists $\mathcal{V}_{F} \subset \mathcal{U}_{F}$ finite such that $\ol{F}\subset \bigcup_{V \in \mathcal{V}_F}V$.

Let $\mathcal{V}= \bigcup_{F \in \cf}\mathcal{V}_{F}$, then $\mathcal{V}$ is a precompact open cover of $X$. For any $x \in X$, there exists $N \in \cn(x)$ such that $\bracs{F \in \cf|N \cap G_F}$ is finite. Thus

is finite, and $\mathcal{V}$ is locally finite.

(3) $\Rightarrow$ (4): By Lemma 4.20.7.

(4) $\Rightarrow$ (5): Let $\seqi{V}, \seqi{W}\subset 2^{X}$ be locally finite refinements of $\mathcal{U}$ consisting of precompact open sets such that for each $i \in I$, $\ol{W_i}\subset V_{i}$.

By Urysohn’s lemma, there exists $\seqi{f}\in C_{c}(X; [0, 1])$ such that for each $i \in I$, $f_{i}|_{\ol{W_i}}= 1$ and $\supp{f_i}\subset V_{i}$.

Let $F = \sum_{i \in I}f_{i}$. For each $x \in X$, there exists $N_{x} \in \cn^{o}(x)$ such that $\bracs{i \in I|N_x \cap V_i \ne \emptyset}$ is finite. In which case,

thus $F|_{N_x}\in C(N_{x}; \real)$. By Lemma 4.6.2, $F \in C(X; \real)$.

Since $\seqi{W}$ is an open cover of $X$, $F(x) > 0$ for all $x \in X$. For each $i \in I$, let $g_{i} = f_{i}/F$, then $g_{i} \in C_{c}(X; [0, 1])$ with $\supp{g_i}= \supp{f_i}\subset W_{i}$. For any $x \in X$, there exists $N_{x} \in \cn^{o}(x)$ such that $\bracs{i \in I|N_x \cap W_i \ne \emptyset}$ is finite. In which case, $\bracs{i \in I|0 < g_i|_{N_x}}$ is also finite. Thus $\seqi{g}$ is a $C_{c}$ partition of unity subordinate to $\mathcal{U}$.

(5) $\Rightarrow$ (1): Let $\mathcal{U}$ be an open cover of $X$ and $\seqi{f}\subset C_{c}(X; [0, 1])$ subordinate to $\mathcal{U}$. For each $i \in I$, let $V_{i} = \bracs{f_i > 0}$, then $\seqi{V}$ is a locally finite refinement of $\mathcal{U}$.

(5) $\Rightarrow$ (6): Take $\mathcal{U}= \bracs{X}$.

(6) $\Rightarrow$ (2): Let $\seqi{f}\subset C_{c}(X; [0, 1])$ be a partition of unity. For each $i \in I$, let $V_{i} = \bracs{f_i > 0}$, then $\seqi{V}$ is a locally finite precompact open cover of $\mathcal{U}$.$\square$

Proof. By Proposition 4.20.5, there exists an exhaustion $\seq{U_n}\subset 2^{X}$ of $X$ by precompact open sets. Denote $U_{0} = \emptyset$. For each $n \in \natp$, let $V_{n} = U_{n+1}\setminus \ol{U_{n-1}}$.

Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_{n} \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n}\setminus \ol{U_{n - 2}}= V_{n-1}$. If $n = 1$, then $x \in U_{2}= V_{1}$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_{m} \cap V_{n} \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of Proposition 4.20.8, $X$ is paracompact.$\square$