Cauchy Sequences

Let $(X,d)$ be a Metric Space. A sequence $\{p_n\}$ in $X$ is called a Cauchy Sequence if

$$\forall \epsilon >0,\exists N\in \mathbb{N},\forall n,m\ge N,d(p_n,p_m)<\epsilon$$

Proposition (Convergence is Cauchy)

Let $\{p_n\}$ be a convergence sequence. Then $\{p_n\}$ is Cauchy.

Proof:

Suppose $p_n\to p$. Let $\epsilon >0$. Since $p_n\to p,\exists N\in \mathbb{N}$. such that $\forall n\ge N,d(p_n,p)<\epsilon /2$. Let $m,n\ge N$. Then

$$\begin{array}{rl}d(p_n,p_m)& \le d(p_n,p)+d(p,q_n)\\ & <\epsilon /2+\epsilon /2\\ & =\epsilon \end{array}$$

And so we are done.

Definition (Completeness)

A Metric Space $(X,d)$ is called complete if every Cauchy Sequence in $X$ also converges in $X$

• $\mathbb{Q}$ is not complete. Consider $p_n\in \mathbb{Q}$ where $\sqrt{2}-1/n<p_n<\sqrt{2}$. We see $p_n\to \sqrt{2}\notin \mathbb{Q}$ so it does not converge in $\mathbb{Q}$, but it is Cauchy.

Lemma (Cauchy is Bounded)

Let $(X,d)$ be a Metric Space. Every Cauchy Sequence is bounded.

Lemma (Cauchy Subsequence Convergence)

Let $\{p_n\}$ be a Cauchy Sequence. Suppose $\exists$ a subsequence $\{p_{n_i}\}$ of $\{p_n\}$ that converges to $p$. Then $p_n\to p$.

Theorem (Compact in Complete)

1. Let $(X,d)$ be a compact metric space. Then $X$ is complete.
2. ${\mathbb{R}}^d$ with the standard metric is complete.

Proof:

1. Let $\{p_n\}$ be a Cauchy Sequence in a compact metric space $X$. Since $X$ is compact, $\exists$ a subsequence $\{p_{n_i}\}$ such that $\{p_{n_i}\}$ converges by Lemma (Cauchy Subsequence Convergence).
2. Let $\{p_n\}$ be a Cauchy Sequence in ${\mathbb{R}}^d$. Then by Lemma (Cauchy is Bounded), $\{p_n\}$ is bounded. This implies that

$$\{p_n\mid n\in \mathbb{N}\}\subset \overline{B_m(0)}$$

for some $m>0$. By Heine-Borel Theorem, $\overline{B_m(0)}$ is compact so by $(1)$, $\{p_n\}$ is convergent. Hence ${\mathbb{R}}^d$ is complete.