> [!definitionb] Definition > > Let $(M, d)$ be a [[Metric Space|metric space]]. $M$ is **complete** if the [[Limit|limit]] of every [[Cauchy Sequence|Cauchy sequence]] in $M$ is also in $M$. > [!theorem] > > Complete metric spaces are cool :) > > *Proof by Staring*. Obvious. > [!theorem] > > Let $(X, d)$ be a metric space and let $X^\prime$ be the set of Cauchy sequences in $X$. For $(x_n), (y_n) \in X^\prime$, define > $ > d^\prime((x_n), (y_n)) = \limv{n} d(x_n, y_n) > $ > Then $d^\prime$ is a pseudometric on $X^\prime$. > > *Proof*. First, if $(x_n) = (y_n)$, then $d(x_n, y_n) = 0$ for all $n \in \nat$ and $\limv{n}d(x_n, y_n) = 0$. Therefore $d^\prime((x_n), (y_n)) = 0$. > > Using the symmetry of $d$, $d(x_n, y_n) = d(y_n, x_n)$ means that > $ > d^\prime((x_n), (y_n)) = \limv{n}d(x_n, y_n) = \limv{n}d(y_n, x_n) = d^\prime((y_n), (x_n)) > $ > > Let $(x_n), (y_n), (z_n) \in X^\prime$, then > $ > d(x_n, z_n) \le d(x_n, y_n) + d(y_n, z_n) > $ > and by the [[Order Limit Theorem|order limit theorem]] > $ > \limv{n}d(x_n, z_n) \le \limv{n}d(x_n, y_n) + \limv{n}d(y_n, z_n) > $ > which means > $ > d^\prime((x_n), (z_n)) \le d^\prime((x_n), (y_n)) + d^\prime((y_n), (z_n)) > $ > [!theorem] > > Let $(x_n), (y_n) \in X^\prime$. Then $(x_n) \sim (y_n)$ if $d^\prime((x_n), (y_n)) = 0$. This is an [[Equivalence Relation|equivalence relation]]. Let $\hat X$ be the set of [[Equivalence Class|equivalence classes]] and define $\hat d$ to be the $d^\prime$ distance between the representatives of the classes. Then $\hat d$ is a well-defined metric on $\hat X$. > > *Proof*. Since $d^\prime((x_n), (x_n)) = 0$, $(x_n) \sim (x_n)$. > > As $d^\prime((x_n), (y_n)) = d^\prime((y_n), (x_n))$, > $ > d^\prime((x_n), (y_n)) = 0 \Leftrightarrow d^\prime((y_n), (x_n)) = 0 > $ > $(x_n) \sim (y_n) \Leftrightarrow (y_n) \sim (x_n)$, the relation is symmetric. > > If $(x_n) \sim (y_n)$ and $(y_n) \sim (z_n)$, then > $ > 0 \le d^\prime((x_n), (z_n)) \le d^\prime((x_n), (y_n)) + d^\prime((y_n), (z_n)) = 0 > $ > $d^\prime((x_n), (z_n)) = 0$ and $(x_n) \sim (z_n)$, the relation is transitive. Therefore $~$ is an equivalence relation. > > Let $x, y \in X^\prime$ and $[x], [y]$ be their respective equivalence classes. Define $\hat d([x], [y]) = d^\prime(x, y)$. > > We first show that this is a well-defined function. Let $[x], [y]$ be equivalence classes and let $x_1, x_2 \in [x]$, $y_1, y_2 \in [y]$. Then > $ > d^\prime(x_1, y_2) \le d^\prime(x_1, y_1) + d^\prime(y_1, y_2) = d^\prime(x_1, y_1) > $ > and > $ > d^\prime(x_1, y_1) \le d^\prime(x_1, y_2) + d^\prime(y_1, y_2) = d^\prime(x_1, y_2) > $ > meaning that > $ > d(x_1, y_1) = d(x_1, y_2) > $ > Applying the same idea to varying $x_1$ and $x_2$ we have > $ > \begin{align*} > d^\prime(x_1, y_1) &\le d^\prime(x_2, y_1) + d^\prime(x_1, x_2) = d^\prime(x_2, y_1) \\ > d^\prime(x_2, y_1) &\le d^\prime(x_1, y_1) + d^\prime(x_1, x_2) = d^\prime(x_1, y_1) \\ > d^\prime(x_1, y_1) &= d^\prime(x_2, y_1) > \end{align*} > $ > Therefore $\hat d([x], [y])$ is well-defined. > > Since $[x]$ is an equivalence class, for any $x_1, x_2 \in [x]$, $d^\prime(x_1, x_2) = 0$ and therefore $\hat d([x], [x]) = 0$. > > Suppose that $\hat d ([x], [y]) = 0$, then > $ > \exists x \in [x], y \in [y]: d^\prime(x, y) = 0 \Rightarrow x \sim y \Rightarrow [x] = [y] > $ > > Using the symmetry of $d^\prime$, > $ > \hat d([x], [y]) = d^\prime(x, y) = d^\prime(y, x) = \hat d([y], [x]) > $ > $\hat d$ is also symmetric. > > Let $[x], [y], [z] \in \hat X$, then > $ > \hat d([x], [z]) = d^\prime(x, z) \le d^\prime(x, y) + d^\prime(y, z) = \hat d([x], [y]) + \hat d([y], [z]) > $ > $\hat d$ satisfies the triangle inequality. > > Therefore $\hat d$ is a metric on $\hat X$. > [!theoremb] Theorem > > The [[Metric Space|metric space]] $(\hat X, \hat d)$ is complete. > > *Proof*. Let $([(x_n)]_m)$ be a Cauchy sequence in $\hat X$ and let $((x_n)_m)$ be a Cauchy sequence of Cauchy sequences in $X$ (a sequence of representatives). Denote the $n$-th element of the $m$-th sequence as $x_{n, m}$. > > Since each $(x_n)_m$ is a Cauchy sequence, > $ > \forall \varepsilon > 0, \exists N_m \in \nat: d(x_{a,m}, x_{b,m}) < \varepsilon \forall a, b \ge N_m > $ > Let $k \in \nat$, $\varepsilon = \frac{1}{5k}$, then > $ > \exists N_{m,k} \in \nat: d(x_{a, m}, x_{b, m}) < \varepsilon \forall a, b \ge N_{m,k} > $ > and take $n_{k,m}$ such that $n_{k,m} \ge N_{m,k}$ and $n_{k,m} > n_{k - 1,m}$. Then $(x_{n_{k,m}})_m$ is a subsequence of $(x_n)_m$ that satisfies > $ > d(x_{n_a, m}, x_{n_b, m}) < \frac{1}{5k} \quad \forall k > 0,\ a, b \ge k > $ > which is also Cauchy. > > As $((x_{n_{k,m}})_m)$ is a Cauchy sequence of Cauchy sequences, > $ > \forall \varepsilon > 0, \exists N_{out} \in \nat: d^\prime((x_{n_{k,a}})_a, (x_{n_{k,b}})_b) < \varepsilon \forall a, b \ge N_{out} > $ > Similarly, let $l \in \nat$, $\varepsilon = \frac{1}{5l}$, then > $ > \exists N_{l} \in \nat: d^\prime((x_{n_{k,a}})_a, (x_{n_{k,b}})_b) < \frac{1}{5l} \forall a, b \ge N_{l} > $ > Take $m_l$ such that $m_l \ge N_l$ and $m_{l} > m_{l - 1}$. Then update the notation because there are so many subscripts, and > $ > ((x_{n_{k, m_l}})_{m_l}) = ((x[n_{k, m_l}])_{m_l}) > $ > is a Cauchy subsequence that satisfies > $ > d^\prime((x[n_{k, m_a}])_{m_a}, (x[n_{k, m_b}])_{m_b}) < \frac{1}{5l} \quad \forall l > 0,\ a, b \ge l > $ > Since $d^\prime((x[n_{k, m_a}])_{m_a}, (x[n_{k, m_b}])_{m_b}) = \limv{n}d(x[n_{k, m_a}, m_a], x[n_{k, m_b}, m_b])$ comes from a Cauchy sequence of real numbers. But as > $ > d(x[n_{g, m_a}, m_a], x[n_{h, m_a}, m_a]) < \frac{1}{5k} \forall g, h \ge k > $ > we have > $ > \limv{g}d(x[n_{g, m_a}, m_a], x[n_{h, m_a}, m_a]) \le \frac{1}{5k} \forall h \ge k > $ > Then if $a, b \ge l$ > $ > \begin{align*} > d(x[n_{k, m_a}, m_a], x[n_{k, m_b}, m_b]) &\le \limv{h}d(x[n_{k, m_a}, m_a], x[n_{h, m_a}, m_a]) \\ > &+\limv{h}d(x[n_{k, m_b}, m_b], x[n_{h, m_b}, m_b]) \\ > &+ \limv{h}d(x[n_{h, m_a}, m_a], x[n_{h, m_b}, m_b]) \\ > &\le \frac{1}{5k} + \frac{1}{5k} \\ > &+ d^\prime((x[n_{k, m_a}])_{m_a}, (x[n_{k, m_b}])_{m_b}) \\ > &< \frac{2}{2k} + \frac{1}{5l} > \end{align*} > $ > > Let $l = k$, and consider the sequence $(x[n_{k, m_k}, m_k])$ (with all terms depending on $k$). Then for any $a, b > k$ > $ > \begin{align*} > d(x[n_{a, m_a}, m_a], x[n_{b, m_b}, m_b]) &\le > d(x[n_{k, m_a}, m_a], x[n_{a, m_a}, m_a]) \\ > &+ d(x[n_{k, m_b}, m_b], x[n_{b, m_b}, m_b]) \\ > &+ d(x[n_{k, m_a}, m_a], x[n_{k, m_b}, m_b]) \\ > &< \frac{1}{5k} + \frac{1}{5k} + \frac{2}{5k} + \frac{1}{5l} \\ > &= \frac{1}{k} > \end{align*} > $ > and for any $\varepsilon > 0$, $\exists N > \frac{1}{\varepsilon}$ such that for any $k \ge N$, > $ > d(x[n_{a, m_a}, m_a], x[n_{b, m_b}, m_b]) < \varepsilon > $ > and therefore $(x[n_{k, m_k}, m_k])$ is a Cauchy sequence. > > Lastly, we show that > $ > \limv{y}d^\prime((x[n_{k, m_k}, m_k]), (x_n)_y) = 0 > $ > Fix $\varepsilon > 0$, then for $k > \frac{4}{\varepsilon}$, > $ > d(x[n_{a, m_a}, m_a], x[n_{b, m_b}, m_b]) < \varepsilon/4 \quad \forall a, b \ge k > $ > We have > $ > d^\prime((x_n)_g, (x_n)_h) < \varepsilon/4 \quad \forall g, h \ge m_k > $ > Then since each $(x_n)_y$ is also a Cauchy sequence, > $ > \begin{align*} > d(x[n_{a, m_a}, m_k], x[n_{b, m_y}, y]) &\le > d^\prime((x[n_{a, m_a}, m_k]), (x_n)_y) \forall y \ge m_k, a \ge k\\ > &+ \frac{\varepsilon}{4} + \frac{\varepsilon}{4} < \varepsilon > \end{align*} > $ > Since > $ > \forall \varepsilon > 0, \exists m_k \in \nat: > d^\prime((x[n_{k, m_k}, m_k]), (x_n)_y)\ > \forall y \ge m_k, a \ge k > $ > we have > $ > \limv{y}d^\prime((x[n_{k, m_k}, m_k]), (x_n)_y) = 0 > $ > and $(x[n_{k, m_k}, m_k]) \in \hat X$ is a Cauchy sequence that represents the sequence of Cauchy sequences.