> [!definition] > > Let $X$ be a [[Set|set]] of points and > $ > d: X \times X \to [0, \infty) > $ > $(X, d)$ is a metric [[Space|space]] if the *metric* [[Function|function]] $d$ satisfies the following axioms for all $x, y \in X$: > - $d(x, y) \ge 0$ and $d(x, y) = 0 \Leftrightarrow x = y$ > - $d(x, y) = d(y, x)$ > - $d(x, z) \le d(x, y) + d(y, z)$ ([[Triangle Inequality]]) > [!definition] > > Let $X = \real^n$ be a [[Normed Vector Space|normed vector space]] in the [[Real Numbers|real numbers]]. Define > $ > d(x, y) = |x - y| > $ > then $d$ is a metric. > [!theorem] > > Let $\phi: [0, \infty] \to [0, 1]$ by > $ > \phi(t) = \begin{cases} > \frac{t}{t + 1} & t < \infty \\ > 1 &t = \infty > \end{cases} > $ > then: > 1. $\phi$ is strictly increasing with $\phi(t + s) \le \phi(t) + \phi(s)$. > 2. If $(Y, \rho)$ is a metric space, then $\phi \circ \rho$ is a bounded metric on $Y$ that defines the same topology as $\rho$. > 3. If $X$ is a [[Topological Space|topological space]], then $\rho(f, g) = \phi(\sup_{x \in X}|f(x) - g(x)|)$ is a metric on $\complex^X$, whose associated topology is the topology of [[Uniform Convergence|uniform convergence]]. > 4. If $X$ is a $\sigma$-compact [[Locally Compact Hausdorff Space|LCH]] space and $\seq{U_n}$ is a [[Sequence|sequence]] of precompact open sets such that $\ol{U_n} \subset U_{n + 1}$, then the function $\rho(f, g) = \sum_{n \in \nat}2^{-n}\phi\paren{\sup_{x \in \ol{U_n}}|f(x) - g(x)|}$ is a metric on $\complex^X$ whose associated topology is the topology of [[Uniform Convergence on Compact Sets|uniform convergence on compact sets]]. > > *Proof*. > > # Inequalities > > Let $t \in [0, \infty)$ and $\varepsilon > 0$, then > $ > \begin{align*} > \phi(t + \varepsilon) - \phi(t) &= \frac{t + \varepsilon}{1 + t + \varepsilon} - \frac{t}{1 + t} \\ > &= \frac{(t + \varepsilon)(1 + t) - t(1 + t + \varepsilon)} > {(1 + t + \varepsilon)(1 + t)} \\ > &= \frac{t + \varepsilon + t^2 + t\varepsilon -t - t^2 - t\varepsilon}{(1 + t + \varepsilon)(1 + t)} \\ > &= \frac{\varepsilon}{(1 + t + \varepsilon)(1 + t)} > 0 > \end{align*} > $ > and > $ > \begin{align*} > \phi(t) + \phi(\varepsilon) &= \frac{t}{1 + t} + \frac{\varepsilon}{1 + \varepsilon} \\ > &\ge \frac{t}{1 + t} + \frac{\varepsilon}{(1 + t + \varepsilon)(1 + t)} \\ > &= \phi(t + \varepsilon) > \end{align*} > $ > > # Equivalent Metric > > If $(Y, \rho)$ is a metric space, then $\phi \circ \rho$ is a bounded metric on $Y$ that defines the same topology as $\rho$. > > *Proof*. Firstly $\phi \circ \rho(x, y) = 0$ if and only if $\rho(x, y) = 0$, if and only if $x = y$. Now, > $ > \begin{align*} > \phi(\rho(x, z)) &\le \phi(\rho(x, y) + \rho(y, z)) \le \phi(\rho(x, y)) + \phi(\rho(y, z)) > \end{align*} > $ > and $\phi \circ \rho$ is a metric. > > Let $x \in X$ and $r > 0$, then since $\phi$ is strictly increasing, > $ > \begin{align*} > B_{\rho}(x, r) &= \bracs{y \in X: \rho(x, y) < r} \\ > &= \bracs{y \in X: \phi \circ \rho (x, y) < \phi(r)} \\ > &= B_{\phi \circ \rho}(x, \phi(r)) > \end{align*} > $ > Since $\phi \circ \rho$ generates the open balls that come from $\rho$, $\topo_{\phi \circ \rho} \supset \topo_{\rho}$. > > Now, for any $r \in (0, 1)$, $B_{\phi \circ \rho}(x, r) = B_{\rho}(x, \phi^{-1}(r))$, where the inverse is possible since $\phi$ is increasing with range $[0, 1]$. Lastly, for any $r \ge 1$, > $ > \begin{align*} > B_{\phi \circ \rho}(x, r) & = \bracs{y \in X: \phi \circ \rho(x, y) < r} \\ > &= \bracs{y \in X: \phi \circ \rho(x, y) < \infty} \\ > &= X > \end{align*} > $ > Since $\rho$ generates the open balls that come from $\phi \circ \rho$, $\topo_{\phi \circ \rho} \subset \topo_\rho$. > > > # Uniform Convergence > > If $X$ is a [[Topological Space|topological space]], then $\rho(f, g) = \phi(\sup_{x \in X}|f(x) - g(x)|)$ is a metric on $\complex^X$, whose associated topology is the topology of [[Uniform Convergence|uniform convergence]]. > > *Proof*. Since $\rho = \sup_{x \in X}|f(x) - g(x)|$ is the metric associated with uniform convergence, $\phi \circ \rho$ is a metric inducing the same topology. > > > ### Uniform Convergence on Compact Sets > > If $X$ is a $\sigma$-compact [[Locally Compact Hausdorff Space|LCH]] space and $\seq{U_n}$ is a [[Sequence|sequence]] of precompact open sets such that $\ol{U_n} \subset U_{n + 1}$, then the function > $ > \rho(f, g) = \sum_{n \in \nat}2^{-n}\phi\paren{\sup_{x \in \ol{U_n}}|f(x) - g(x)|} > $ > is a metric on $\complex^X$ whose associated topology is the topology of [[Uniform Convergence on Compact Sets|uniform convergence on compact sets]]. > > ### Metric > > $\rho$ is a metric on $\complex^X$. > > *Proof*. Denote > $ > \rho_n(f, g) = \sup_{x \in \ol{U_n}}|f(x) - g(x)| > $ > Let $f, g, h \in \complex^X$, then for all $n \in \nat$, > $ > \rho_n(f, h) \le \rho_n(f, g) + \rho_n(g, h) > $ > therefore > $ > \begin{align*} > \rho(f, h) &=\sum_{n \in \nat}2^{-n}\phi\paren{\rho_n(f, h)} \\ > &\le \sum_{n \in \nat}2^{-n}\phi\paren{\rho_n(f, g) + \rho_n(g, h)} \\ > &\le \sum_{n \in \nat}2^{-n}\braks{\phi \circ \rho_n(f, g) + \phi \circ \rho_n(g, h)} \\ > &\le \rho(f, g) + \rho(g, h) > \end{align*} > $ > > Now, let $f, g \in \complex^X$. If $f - g \ne 0$, then there exists $x \in X$ such that $f(x) - g(x) \ne 0$, and $n \in \nat$ such that $x \in U_n$, in which case > $ > \rho(f, g) \ge 2^{-n}\phi\paren{\rho_n(f, g)} > 0 > $ > > Lastly, let $f, g \in \complex^X$, then > $ > \begin{align*} > \rho(f, g) &= \sum_{n \in \nat}2^{-n}\phi(p_n(f,g)) \\ > &= \sum_{n \in \nat}2^{-n}\phi(p_n(g, f)) \\ > &= \rho(g, f) > \end{align*} > $ > Therefore $\rho$ is a metric. > > > ### Topology > > $\rho$ induces the topology of uniform convergence on compact sets. > > ##### Notation > > Let $\topo_U$ be the topology of uniform convergence, and $\topo_\rho$ be the topology induced by $\rho$. Denote $B(f, \varepsilon) = \bracs{g \in \complex^X: \rho(f, g) < \varepsilon}$, and > $ > B(f, n, \varepsilon) = \bracs{g \in \complex^X: \rho_n(f, g) < \varepsilon} > $ > then > $ > \ce = \bracs{B(f, n, \varepsilon): f \in \complex^X, n \in \nat, \varepsilon > 0} > $ > generates the topology of uniform convergence on compact sets. > > > ##### Metric Generates Uniform Convergence > > Let $f \in \complex^X$, $n \in \nat$ and $r > 0$, then $B(f, n, r) \in \topo_\rho$. Therefore $\topo_U \subset \topo_\rho$. > > *Proof*. Let $g \in B(f, n, r)$, then $\rho_n(f, g) < r$. Let $\varepsilon_g = r - \rho_n(f, g)$, then for any $h \in B(g, n, \varepsilon_g)$, $\rho_n(f, h) \le \rho_n(f, g) + \rho_n(g, h) < \varepsilon_g$, and $h \in B(f, n, r)$. > > Since $\rho(g, h) = \sum_{n \in \nat}2^{-n}\rho_n(g, h)$, if $\rho(g, h) < 2^{-n}\phi(\varepsilon_g)$, then > $ > \begin{align*} > 2^{-n}\phi(\rho_n(g, h)) &\le \sum_{k \in \nat}2^{-k}\phi(\rho_n(g, h)) \le 2^{-n}\phi(\varepsilon_g) \\ > 2^{-n}\phi(\rho_n(g, h)) &\le 2^{-n}\phi(\varepsilon_g) \\ > \rho_n(g, h) &\le \varepsilon_g > \end{align*} > $ > and we can write > $ > B(f, n, r) = \bigcup_{g \in B(f, n, r)}B(g, 2^{-n}\phi(\varepsilon_g)) > $ > as a union of open balls. > > > ##### Uniform Convergence Generates Metric > > Let $f \in \complex^X$ and $r > 0$, then $B(f, r) \in \topo_U$. Therefore $T_\rho \subset T_U$. > > *Proof*. Let $g \in B(f, r)$, and $\varepsilon_g > 0$ such that $B(g, \varepsilon_g) \subset B(f, r)$. Choose $n_g \in \nat$ such that $2^{-n_g} < \varepsilon_g / 2$. Let $\delta_{g}$ be such that > $ > \sum_{k = 1}^{n_g}2^{-k}\phi(\delta_{g}) < \varepsilon_g/2 > $ > Then whenever $\rho_{n_g}(g, h) < \delta_g$, as $\rho_m(g, h) \le \rho_n(g, h)$ whenever $n \ge m$, > $ > \begin{align*} > \sum_{n \in \nat}2^{-n}\phi(\rho_n(g, h)) &= \sum_{k = 1}^{n_g}2^{-k}\phi(\rho_k(g, h)) + \sum_{k > n_g}2^{-k}\phi(\rho_k(g, h)) \\ > &< \underbrace{\sum_{k = 1}^{n_g}2^{-k}\phi(\delta_{g})}_{< \varepsilon_g / 2} + \underbrace{\sum_{k > n_g}2^{-k}\phi(\rho_k(g, h))}_{< \varepsilon_g / 2} \\ > &< \varepsilon_g > \end{align*} > $ > and we have found $B(g, n_g, \delta_{g}) \subset B(g, \varepsilon_g)$, meaning that > $ > B(f, r) = \bigcup_{g \in B(f, r)}B(g, n_g, \delta_g) > $ > is a union of "open balls". > [!theorem] > > Let $U$ be an open set and $V \subset \subset U$ be a compactly contained open set, then > $ > d(V, \partial U) > 0 > $ > *Proof*. Let $K = \ol{V}$, then $K \cap \partial U = \emptyset$. If $d(K, \partial U) = 0$, then there exists $\seq{x_n} \subset \partial U$ and $\seq{y_n} \subset K$ such that $d(x_n, y_n) \to 0$. By sequential compactness, there exists $y \in K$ such that $d(x_n, y) = 0$. This implies that $K \cap \partial U \ne \emptyset$.