Completeness Criterion (Metric Spaces)/Proof 1

Theorem
Let $\left({X, d}\right)$ be a metric space.

Let $A \subseteq X$ be a dense subset.

Suppose that every Cauchy sequence in $A$ converges in $X$.

Then $X$ is complete.

Proof
Let $\left\langle{x_n}\right\rangle_{n \in \N}$ be a Cauchy sequence in $X$.

For each $n$ pick a Cauchy sequence $\left\langle{y_{n,m}}\right\rangle_{m \in \N}$ in $A$ converging to $x_n$ like so:


 * CompletenessCriterionProof.png

Let $N \in \N$ be such that $d \left({x_{n_1}, x_{n_2}}\right) < \epsilon / 3$ for all $n_1, n_2 > N$.

Let $M \in \N$ be such that $d \left({y_{n_i, m}, x_{n_i}}\right) < \epsilon / 3$ for all $m > M$ and all $n_1, n_2 > N$.

Let $m > M$.

Let $n_1, n_2 > N$.

We have:

Therefore $\left\langle{y_{m,n}}\right\rangle_{n \in \N}$ is Cauchy in $A$ for $m > M$.

So $\left\langle{y_{m,n}}\right\rangle_{n \in \N}$ converges to some limit $y_n \in X$.