> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is: > 1. [[Locally Compact]] if every point has a [[Compactness|compact]] [[Neighbourhood|neighbourhood]]. > 2. [[Hausdorff Space|Hausdorff]] if points can be separated by disjoint [[Open Set|open sets]]. > 3. [$\sigma$-compact](Sigma%20Compact) if it is a countable union of compact sets. > > If $X$ is all three, then $X$ is a $\sigma$-compact [[Locally Compact Hausdorff Space|LCH]] space. > [!theorem] > > Let $X$ be a [[Separable Topological Space|separable]] [[Locally Compact Hausdorff Space|LCH]] space, then $X$ is $\sigma$-compact. > > *Proof*. Let $\seq{x_n} \subset X$ be a countable dense subset, and let $K_n \in \cn^k(x)$, then $K_n$ is compact and $X = \bigcup_{n \in \nat}K_n$. > [!theorem] > > Let $X$ be a [$\sigma$-compact](Sigma%20Compact) LCH space, then there is a [[Sequence|sequence]] $\seq{U_i}$ of [[Precompact|precompact]] [[Open Set|open sets]] such that $\ol{U_n} \subset U_{n + 1}$ and $X = \bigcup_{i \in \nat}U_i$. > > *Proof*. Let $X = \bigcup_{n \in \nat}K_n$ be where each $K_n$ is compact. > > Let $U_1 \in \cn(K_1)^o$ be a precompact open neighbourhood of $K_1$. For any $n \in \nat$, take a precompact open neighbourhood, > $ > U_n \in \cn\paren{K_n \cup \bigcup_{i = 1}^{n - 1}\ol{U_i}}^o > $ > then each $U_n \supset \bigcup_{i = 1}^{n - 1}\ol{U_i} = \ol{U_{n - 1}}$ and $\bigcup_{n \in \nat}U_n = X$. > [!theorem] > > Let $X$ be a $\sigma$-compact LCH space and $\seq{U_n}$ be a sequence of precompact open sets such that $\ol{U_n} \subset U_{n + 1}$ for all $n$ and $\bigcup_{n \in \nat}U_n = X$, then > $ > \ce = \bracs{ > \bracs{g \in \complex^X: \sup_{x \in \ol{U_m}}|f(x) - g(x)| < \frac{1}{n}} > \Bigg|m, n \in \nat > } > $ > is a [[Neighbourhood Base|neighbourhood base]] for $f \in \complex^X$ in the topology of [[Uniform Convergence on Compact Sets|uniform convergence on compact sets]]. > > Hence, this topology is [[First Countable|first countable]] with $f_n \to f$ if and only if $f_n \to f$ [[Uniform Convergence|uniformly]] on every $\ol{U_n}$. > > > *Proof*. > > ### Open Neighbourhood Reduction > > Denote the generating sets of the topology as > $ > B_{K}(f, n) = \bracs{g \in \complex^X: \sup_{x \in K}\abs{g(x) - f(x)} < \frac{1}{n}} > $ > then for any $U \in \cn(f)^o$, there exists $K$ compact and $n \in \nat$ such that $f \in B_K(f, n) \subset U$. > > *Proof*. Suppose that $B_K(g, n) \in \cn(f)^o$ is an open neighbourhood of $f$, then there exists $m \in \nat$ such that $f \in B_K(f, m) \subset B_K(g, n)$[^2] and we can assume that neighbourhoods in the generating set are "centred" at $f$. > > Let $U \in \cn(f)^o$ be an open neighbourhood of $f$, then it contains a finite intersection of of the generating sets $\bigcap_{i = 1}^{n}B_{K_i}(g_i, n_i)$, where each $K_i$ is compact, and $B_{K_i}(g_i, n_i)$ is an open neighbourhood of $f$. Therefore we can write > $ > f \in \bigcap_{i = 1}^{n}B_{K_i}(f, m_i) \subset \bigcap_{i = 1}^{n}B_{K_i}(g_i, n_i) > $ > where taking $K = \bigcup_{i = 1}^{n}K_i$ and $m = \max_{i = 1}^{n}m_i$ yields > $ > \bigcap_{i = 1}^{n}B_{K_i}(f, m_i) \supset \bigcap_{i = 1}^{n}B_{K_i}(f, m) = B_{K}(f, m) > $ > and $K$ is compact since it is a finite union of compact sets. > > Therefore any open neighbourhood $U$ can be reduced to $B_{K}(f, m)$ with $f \in B_{K}(f, m) \subset U$ where $m \in \nat$ and $K$ compact. > > ### Compactness Reduction > > Let $\seq{U_n}$ be a sequence of precompact open sets such that $\ol{U_n} \subset U_{n + 1}$ for all $n \in \nat$ and $\bigcup_{n \in \nat}U_n = X$. Then for any compact set $K$ there exists $j \in \nat$ such that $U_j \supset K$. > > *Proof*. Since $\bigcup_{n \in \nat}U_n = X$, $\seq{U_n}$ forms an [[Open Cover|open cover]] of $K$, there exists a finite subcover $\bracs{U_j}_{j \in J}$ with $J \subset \nat$ finite. Let $j = \max(J)$, as $U_j \supset U_{k}$ for all $j \le j$, $U_j \supset K$. Therefore for any $K$ compact there exists > > ### Neighbourhood Base > > The collection > $ > \ce(f) = \bracs{B_{K}(f, n): n \in \nat} > $ > forms a neighbourhood base for $\topo$ at $f$. > > *Proof*. Let $U \in \cn(f)^o$, then there exists $B_K(f, n)$ such that $f \in B_K(f, n) \subset U$. Since $K$ is compact, there exists precompact $U_j$ such that $U_j \supset K$. Take $B_{\ol{U_j}}(f, n) \in \ce(f)$, then > $ > f \in B_{\ol{U_j}}(f, n) \subset B_K(f, n) \subset U > $ > Therefore $\ce(f)$ forms a neighbourhood base at $f$. > [!theorem] > > Let $(X, \topo)$ be a [$\sigma$-compact](Sigma%20Compact) [[Locally Compact Hausdorff Space|LCH]] space, and $\mathcal{U}$ be an [[Open Cover|open cover]], then there exists locally finite refinements $\mathcal V$ and $\mathcal W$ such that $\ol{V}$ is compact for all $V \in \mathcal V$, and $\ol W$ is contained in a set in $\mathcal V$. > > *Proof*. Let $\seq{U_n}$ be a [[Sequence|sequence]] of precompact open sets such that $\ol{U_n} \subset U_{n + 1}$ and $\bigcup_{n \in \nat}U_n = U$. Let $\mathcal U$ be $n \in \nat$, then > $ > V_n = \bracs{E \cap (U_{n + 2} \setminus \ol{U_{n - 1}}): E \in \mathcal{U}} > $ > is an open cover of $\ol{U_{n + 1}} \setminus U_{n}$. As $\ol{U_{n + 1}} \setminus U_{n}$ is compact, $V_n$ has a finite subcover $\mathcal{V}_n$. Let $\mathcal{V} = \bigcup_{n \in \nat}\mathcal{V}_n$, then $\mathcal{V}$ is a refinement of $\mathcal{U}$. For any point $x \in U_n \setminus \ol{U_{n - 1}}$, only sets in $\mathcal{V}_n, \mathcal{V}_{n - 1}, \mathcal{V}_{n - 2}$ can intersect $U_n \setminus U_{n - 1}$. Therefore $\mathcal{V}$ is a locally finite refinement consisting of precompact open sets. > > Let $x \in X$ and $V \in \mathcal V$, then there exists a precompact open neighbourhood $W_x$ with $\ol{W_x} \subseteq V$. Let $W_V = \bracs{W_x: x \in V}$, then it has a finite subcover $\mathcal{W}_V$ of $\ol{V}$. Let $\mathcal{W} = \bigcup_{V \in \mathcal V}\mathcal{W}_V$, and $x \in X$ with $x \in U_n \setminus \ol{U_{n - 1}}$. Since only finitely many $V$s intersect $U_n \setminus \ol{U_{n - 1}}$, the only sets in $\mathcal{W}$ that intersect $U_n \setminus \ol{U_{n - 1}}$ are contained in $\bigcup_{V \cap U_{n} \setminus \ol{U_{n - 1}} \ne \emptyset}\mathcal{W}_V$. > [!theorem] > > Let $(X, \topo)$ be a $\sigma$-compact LCH space, and $\mathcal{U}$ be an [[Open Cover|open cover]], then there exists a [[Partition of Unity|partition of unity]] $E \subset C(X, [0, 1])$ subordinate to $\mathcal{U}$. > > *Proof*. Let $\seq{U_n}$ be a sequence of precompact open sets such that $X = \bigcup_{n \in \nat}U_n$ and $\ol{U_n} \subset U_{n + 1}$ for all $n \in \nat$, then $K_n = \ol{U_{n}} \setminus U_{n - 1}$ ($U_0 = \emptyset$) is compact. Since $X$ is LCH, each $K_n$ has a finite partition of unity $\bracs{f_{(n, k)}}$ that is compactly supported, and we can choose them to be supported within $\ol{U_{n + 1}} \setminus U_{n - 2}$. This way, combining the partitions for each $K_n$ yields a set of functions whose sum is finite on each point. Normalising them gives the desired partition.