> [!definition] > > A [[Sequence|sequence]] $(x_n)$ is a *Cauchy sequence* if $\forall \varepsilon > 0$, $\exists N \in \nat$, where whenever $m, b \ge N$, $|x_{n} - x_m| \lt \varepsilon$. > [!theorem] > > Cauchy sequences are bounded. > > *Proof*. Given $\varepsilon = 1$, $\exists N \in \nat:$ $|x_m - x_n| < 1$, $\forall m, n \ge N$. Then, $|x_n| < |x_N| + 1 \forall n \ge N$. Then, > $ > M = \max\{|x_1|, |x_2|, |x_3| \cdots, |x_{N - 1}|, |x_N| + 1\} > $ > is a bound for the sequence. > [!theorem] > > Every [[Limit|convergent]] sequence is a Cauchy sequence. > > *Proof*. Suppose that $(x_n)$ converges to $x$. Then, $\forall \varepsilon > 0$, $\exists N \in \nat$ where for all $n \ge N$, $|x_n - x| < \frac{\varepsilon}{2}$. Let $m \ge N$, then > $ > \varepsilon = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} > > |x_n - x| + |x - x_m| \ge > |x_n - x + x - x_m| = |x_n - x_m| > $ > $(x_n)$ is a Cauchy sequence. > > Note that this direction is independent from Completeness. > [!theoremb] Cauchy Criterion > > Every Cauchy sequence of [[Real Numbers|real numbers]] is convergent. Otherwise, the Cauchy criterion is used as the definition of convergent sequences. > > *Proof*. Let $(x_n)$ be a Cauchy sequence. Use the [[Bolzano-Weierstrass Property|Bolzano-Weierstrass Theorem]] (based on Completeness) to create a convergent subsequence $(x_{n_k}) \to x$. Let $\varepsilon > 0$, then since $(x_n)$ is Cauchy, > $ > \exists N \in \nat: |x_n - x_m| < \frac{\varepsilon}{2} \quad \forall m, n \ge N > $ > Since $(x_{n_{k}}) \to x$, choose a term in the subsequence $x_{n_{K}}$ with $n_K \ge N$, where > $ > |x_{n_{K}} - x| \lt \frac{\varepsilon}{2} > $ > But then since $n_K \ge N$, > $ > \begin{align*} > |x_n - x| &= |x_n - x_{n_{K}} + x_{n_{k}} - x| \\ > &\le |x_n - x_{n_{K}}| + |x_{n_K} - x| \\ > &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon > \end{align*} > $ > [!theorem] > > Let $(x_n)$ and $(y_n)$ be Cauchy sequences in a [[Metric Space|metric space]] $(X, d)$. Then $(d(x_n, y_n))$ is a Cauchy sequence of real numbers. > > *Proof*. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, > $ > \forall \varepsilon > 0, \exists N_x \in \nat: \forall m, n \ge N_x, d(x_m, x_n) < \varepsilon/2 > $ > and > $ > \forall \varepsilon > 0, \exists N_y \in \nat: \forall m, n \ge N_y, d(x_m, x_n) < \varepsilon/2 > $ > Fix $\varepsilon > 0$ and take $N = \max (N_x, N_y)$, and have $m, n \ge N$. Using the triangle inequality we have > $ > d(x_n, y_n) \le d(x_m, y_m) + d(x_m, x_n) + d(y_m, y_n) > $ > and > $ > d(x_m, y_m) \le d(x_n, y_n) + d(x_m, x_n) + d(y_m, y_n) > $ > which means > $ > |d(x_n, y_n) - d(x_m, y_m)| < d(x_m, x_n) + d(y_m, y_n) > < 2\varepsilon/2 = \varepsilon > $ > and therefore $(d(x_n, y_n))$ is also Cauchy.