Completeness Criterion (Metric Spaces)

Theorem
Let $(X, d)$ 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 1
Let $(x_n)_{n\in\N}$, be a Cauchy sequence in $X$.

For each $n$ pick a Cauchy sequence $(y_{n,m})_{m \in \N}$ in $A$ converging to $x_n$ like so:


 * CompletenessCriterionProof.png

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

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

Now let $m > M$, Let $n_1,n_2 > N$. We have:

Therefore, $(y_{m,n})_{n \in \N}$ is Cauchy in $A$ for $m > M$, and so converges to some limit $y_n \in X$.

Proof 2
Let $(x_n)$ be a Cauchy sequence in $X$.

Since $A$ is dense, we can choose for each $n$ some $y_n \in A$ which is within $1/n$ of $x_n$.

We will show that $(y_n)$ is Cauchy.

Let $\varepsilon > 0$.


 * Since $(x_n)$ is Cauchy, there is some $N'$ such that for each $n, m \geq N'$, $d(x_n, x_m) < \frac{1}{3} \varepsilon$.


 * Let $N \in \mathbb N$ be greater than $N'$ and $3 \varepsilon^{-1}$.


 * Note that $n, m \geq N$ implies that both


 * $n, m \geq N'$, and


 * $1/n, 1/m \leq 1/N < \frac{1}{3} \varepsilon$.


 * Thus, for $n, m \geq N$, we have by the triangle inequality and above comments that


 * $\array{

d(y_n, y_m)   &    \leq    &    d(y_n, x_n) &+& d(x_n, x_m) &+& d(x_m, y_m)    \\ &   <       &    1/n         &+& \frac{1}{3} \varepsilon &+& 1/m    \\ &   <    &    \frac{1}{3} \varepsilon &+& \frac{1}{3} \varepsilon &+& \frac{1}{3} \varepsilon    \\ &   =       &    \varepsilon }$

Hence $(y_n)$ is a Cauchy sequence in $A$, and by the assumption, it converges to some limit $y \in X$.

Now, we verify that $(x_n)$ converges to $y$ as well.

Let $\varepsilon > 0$.


 * Since $(y_n)$ converges, there is some $N'$ such that for each $n \geq N'$, $d(y_n, y) < \frac{1}{2} \varepsilon$.


 * Let $N \in \mathbb N$ be greater than $N'$ and $2 \varepsilon^{-1}$


 * Note that $n \geq N$ implies that both


 * $n \geq N'$, and


 * $1/n \leq 1/N < \frac{1}{2} \varepsilon$.


 * Thus, for $n \geq N$, we have by the triangle inequality and above comments that


 * $\array{

d(x_n, y)   &    \leq    &    d(x_n, y_n) &+& d(y_n, y)    \\ &   <       &    1/n         &+& \frac{1}{2} \varepsilon    \\ &   <       &    \frac{1}{2} \varepsilon  &+& \frac{1}{2} \varepsilon    \\ &   =       &    \varepsilon }$

Hence $(x_n)$ converges to $y$.