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:

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle d(y_{n_1,m},y_{n_2,m})\)	\(\leq\)	\(\displaystyle d(y_{n_1,m},x_{n_1}) + d(x_{n_1},x_{n_2}) + d(x_{n_2},y_{n_2,m})\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	two applications of the triangle inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \epsilon\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $\epsilon > 0$.

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

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

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

$n, m \geq N'$, and

$1/n, 1/m \leq 1/N < \frac 1 3 \epsilon$.

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

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

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 $\epsilon > 0$.

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

$1/n \leq 1/N < \frac 1 2 \epsilon$.

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

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

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

$\blacksquare$

Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.