---
tag: DRH0F7968Y
kind: proof
name: Proof that a metric space is complete if and only if each Cauchy sequence has a convergent subsequence
msc: ["54E50"]
proves: 0N1ZWFJDN5
uses: [063W5JMY85, 08EWT7J0SC, 0MCK8K2D7Z, 06JYPJ4WHH, 0R8J3ZQMJS, 01M6CW9VM6]
sources:
  - url: https://proofwiki.org/wiki/Convergent_Subsequence_of_Cauchy_Sequence
---

## Proof

We prove each implication. The key step shows that a Cauchy sequence follows its convergent subsequences.

1. Let $(X, d)$ be a [[metric space|063W5JMY85]].
2. Assume $X$ is [[complete|08EWT7J0SC]]. Let $(x_n)$ be a [[Cauchy sequence|0MCK8K2D7Z]] in $X$. By completeness, it [[converges|06JYPJ4WHH]] to some $x \in X$.
3. With $n_k = k$, the sequence $(x_{n_k})$ is a [[subsequence|0R8J3ZQMJS]] of $(x_n)$, namely $(x_n)$ itself. So $(x_n)$ has a convergent subsequence.
4. Now assume that each Cauchy sequence in $X$ has a convergent subsequence. Let $(x_n)_{n \ge 1}$ be a Cauchy sequence in $X$.
5. By assumption, there are integers $1 \le n_1 < n_2 < \cdots$ and a point $x \in X$ with $x_{n_k} \to x$ as $k \to \infty$.
6. The $n_k$ are strictly increasing integers with $n_1 \ge 1$. By induction on $k$, we get $n_k \ge k$ for all $k \ge 1$.
7. Let $\varepsilon > 0$ be real. Since $(x_n)$ is Cauchy, there is an integer $N$ with $d(x_m, x_n) < \varepsilon / 2$ for all $m, n \ge N$.
8. Since $x_{n_k} \to x$, there is an integer $L$ with $d(x_{n_k}, x) < \varepsilon / 2$ for all $k \ge L$.
9. Let $k = \max(N, L)$. Then $k \ge L$, and $n_k \ge k \ge N$ by step 6.
10. Let $n \ge N$. By the [[triangle inequality|01M6CW9VM6]] and steps 7 to 9, $d(x_n, x) \le d(x_n, x_{n_k}) + d(x_{n_k}, x) < \varepsilon / 2 + \varepsilon / 2 = \varepsilon$.
11. Since $\varepsilon$ was arbitrary, $x_n \to x$. So each Cauchy sequence in $X$ converges, and $X$ is complete.

## Idea

In a complete space, the whole sequence converges, so it is its own convergent subsequence.
Conversely, late terms of a Cauchy sequence are close to each other, and some late term of the subsequence is close to the limit. The triangle inequality joins the two estimates.