> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is [[Locally Compact|locally compact]] if every point has a [[Compactness|compact]] [[Neighbourhood|neighbourhood]]. Locally compact [[Hausdorff Space|Hausdorff spaces]] can be abbreviated as LCH. > [!theorem] > > Let $X$ be a LCH space and $E \subset X$, then $E$ is [[Closed Set|closed]] if and only if $E \cap K$ is closed for every compact $K \subset X$. > > *Proof*. Since $X$ is Hausdorff, all compact sets are closed and $E \cap K$ is closed for all compact sets $K$. > > If $E$ is not closed, then there exists an [[Accumulation Point|accumulation point]] $x \in \ol{E} \setminus E$ of $E$. Let $K$ be a compact neighbourhood of $x$, then $x$ is an accumulation point of $E \cap K$, not in $E \cap K$. Therefore $E \cap K$ is not closed. # Neighbourhood Properties > [!theorem] > > Let $X$ be a LCH space, $x \in X$ be a point and $U \in \cn(x)^o$ be an [[Open Set|open]] neighbourhood of $x$, then there exists a compact neighbourhood $N \in \cn(x)$ such that $x \in N \subset U$. > > *Proof*. Suppose that $\ol{U}$ is compact, then in the [[Relative Topology|relative topology]] there exists disjoint open sets $V$ and $W$ such that $x \in V$ and $\partial U \subset W$[^1]. This gives $\ol{V} \subset W^c \subset U$ and as $\ol{U}$ is compact, so is $\ol{V}$ in the relative topolology. Since $\ol{U}$ is closed in $X$, $\ol{V}$ is also closed in $X$ and therefore compact. > > If $\ol{U}$ is not compact, take a compact neighbourhood $N \in \cn(x)$ and replace $U$ with $U \cap N^o$. > [!theorem] > > Let $X$ be a LCH space and $K \subset U \subset X$ where $K$ is compact and $U$ is open. Then there exists a [[Precompact|precompact]] $V$ such that $K \subset V \subset \ol{V} \subset U$. > > *Proof*. For any $x \in K$, let $N_x$ be a compact neighbourhood of $x$ such that $x \in N_x \subset U$. Then $\bracs{U_x^o}_{x \in K}$ is an [[Open Cover|open cover]] of $K$, and has a finite subcover $\bracs{U_i}_1^n$. Let $V = \bigcup_{i = 1}^{n}U_i$, then $\ol{V} = \bigcup_{i = 1}^{n}\ol{U_i} \subset U$ is a finite union of compact sets, and therefore compact. # Continuous Functions > [!theorem] [[Urysohn's Lemma]], LCH Version > > Let $X$ be a LCH space and $K \subset U \subset X$ where $K$ is compact and $U$ is open. Then there exists a [[Continuity|continuous]] function $f \in C(X, [0, 1])$ such that $f = 1$ on $K$ and $0$ outside of a compact subset of $U$. > > Therefore all LCH spaces are [[Completely Regular Space|completely regular]]. > > *Proof*. Let $V$ be a precompact open neighbourhood of $K$ with $K \subset V \subset \ol{V} \subset U$. Then $\ol{V}$ is a compact Hausdorff space, and therefore [[Normal Space|normal]]. Since $K$ is compact in $X$, it is also compact in $\ol{V}$ and is closed. > > By [[Urysohn's Lemma]], there exists $f \in C(\ol{V}, [0, 1])$ that is $1$ on $K$ and $0$ on $\partial V$. Extend $f$ to $X$ by setting it to $0$ on $\ol{V}^c$ and denote the extension as $F$. Let $E \subset [0, 1]$ be closed, then > $ > F^{-1}(E) = f^{-1}(E) \cup \paren{F|_{\ol{V}^c}}^{-1}(E) > $ > If $0 \not\in E$, then $F^{-1}(E) = f^{-1}(E)$ is closed. If $0 \in E$, then $F^{-1}(E) = f^{-1}(E) \cup \ol{V}^c$. As $f^{-1}(0) \supset \partial V$, > $ > F^{-1}(E) = f^{-1}(E) \cup \ol{V}^c = f^{-1}(E) \cup V^c > $ > which is closed. Therefore $f$ is continuous. > [!theorem] [[Tietze Extension Theorem]], LCH version > > ![[tietze_lch.png]] > > Let $X$ be a LCH space and $K \subset X$ be compact. If $f \in C(K)$ is a [[Continuity|continuous]] function, then there exists $F \in C(K)$ such that $F|_K = f$. Moreover, $F$ can be taken to vanish outside of a compact set. > > *Proof*. Let $V$ be a precompact open neighbourhood of $K$ such that $K \subset V \subset \ol{V}$, then $\ol{V}$ is compact and normal. As $K$ is compact in a compact Hausdorff space $\ol{V}$, $K$ is closed. > > By the [[Tietze Extension Theorem]], there is an extension $F$ of $f$ in $\ol{V}$. Moreover, in the construction of $\phi$ in the proof, we can take $B' = B \cup \partial V$, which is still closed, and does not influence the approximation. Therefore we can take $F$ to be $0$ outside of $\ol{V}$. > > Similar to Urysohn's Lemma, we can decompose any preimage > $ > F^{-1}(E) = F|_{\ol{V}}^{-1}(E) \cup \ol{V}^c = F|_{\ol{V}}^{-1}(E) \cup V^c > $ > with the $V^c$ being optional. This way, the preimage of every closed set is closed, and the preimage of every open set is open. Therefore the extension is continuous. > [!theorem] > > Let $X$ be a LCH space, then the space $C_0(X)$ of functions that [[Vanishes at Infinity|vanishes at infinity]] is the [[Topological Closure|closure]] of the space $C_c(X)$ of [[Compactly Supported|compactly supported]] functions with respect to the [[Uniform Norm|uniform norm]]. > > *Proof*. Let $\seq{f_n} \subset C_c(X)$ such that $f_n \to f$ [[Uniform Convergence|uniformly]], then for all $\varepsilon > 0$ there exists $n \in \nat$ such that $\norm{f_n - f} < \varepsilon$. As a result, for any $x \not\in \supp{f_n}$, $|f(x)| < \varepsilon$ and $\bracs{x: |f(x)| \ge \varepsilon} \subset \supp{f_n}$ is a closed subset of a compact set, and therefore compact. So $\ol{C_c(X)} \subset C_0(X)$. > > Now let $f \in C_0(X)$ and $n \in \nat$ with $K_n = \bracs{x: |f(x)| \ge 1/n}$, then $K_n$ is compact. By the LCH version of Urysohn's Lemma, there is an extension of $f$ to a continuous function $g_n$ that is $1$ on $K_n$ and $0$ outside of some compact neighbourhood $\ol{V_n}$ of $K_n$. Take $f_n = fg_n$, then $f_n \in C_0(X)$ vanishes outside of $\ol{V_n}$ with > $ > \norm{f_n - f} = \norm{f(1 - g_n)} \le \norm{f \cdot \chi_{K_n^c}} < \frac{1}{n} > $ > so $f_n \to f$ uniformly. > [!theorem] > > Let $X$ be a LCH space, then the [[Space of Continuous Functions|space of continuous functions]] $C(X)$ is a [[Closed Set|closed]] subspace of $\complex^X$ under the topology of [[Uniform Convergence on Compact Sets|uniform convergence on compact sets]]. > > *Proof*. Let $f \in \ol{C(X)}$, then $f$ is an [[Topological Closure|adherent point]] of $C(X)$, and cannot be separated from elements of $C(X)$ with the uniform norm on compact sets. > > Let $K \subset X$ be [[Compactness|compact]], then for any $n \in \nat$, > $ > \exists f_n \in C(X): \sup_{x \in K}|f_n(x) - f(x)| < \frac{1}{n} > $ > meaning that $f_n|_K \to f|_K$ uniformly, and $f$ is continuous on $K$. > > Let $x \in X$ and $K \in \cn(x)$ be a compact neighbourhood, then $f$ is continuous on $K$. Since for any neighbourhood $E$ of $f(x)$, $f^{-1}(E)$ is a neighbourhood of $x$ in $K$, and $f^{-1}(E)^o \cap K^o$ is an open neighbourhood of $x$ in $X$. Therefore $f$ is continuous at $x$. > > As $f$ is continuous at $x$ for all $x \in X$, $f$ is continuous. # Partition of Unity > [!theorem] > > Let $X$ be a LCH space, $K \subset X$ compact, and $\seqf{U_j}$ be an open cover of $K$, then there exists a partition of unity on $K$, subordinate to $\seqf{U_j}$, consisting of [[Compactly Supported|compactly supported]] functions. > > *Proof*. Let $x \in K$, then there exists a compact neighbourhood $N_x \subset U_j$ for some $j \in [n]$. Since $\bracs{N_x^o}_{x \in K}$ forms an open cover of $K$, there exists a finite subcover $\bracs{N_j}_1^k$. Let $F_j$ be the union of $N_i$s that are contained in $U_j$, then $F_j \subset U_j$ is compact. By Urysohn's lemma, there exists $g_j \in C_c(X, [0, 1])$ such that $g_j = 1$ on $F_j$, and $\supp{g_j} \subset U_j$. > > By Urysohn's lemma again, there exists $f \in C_c(X, [0, 1])$ with $f = 1$ on $K$ and $\supp{f} \subset \bigcup_{j = 1}^nU_j$. Let $g' = 1 - f$, then $g' + \sum_{j = 1}^{n}g_j$ is non-zero on $K$. Normalise each term as > $ > h_j = \frac{g_j}{g' + \sum_{k = 1}^ng_k} > $ > then each $h_j$ is still continuous and sum up to $1$ on $K$. [^1]: The [[Boundary|boundary]] is closed since $\partial U = \ol{U} \setminus U^o$. [^2]: See [[Uniform Convergence on Compact Sets]].