Completeness Criterion (Metric Spaces)/Proof 1

Theorem

Let $M = \struct {S, d}$ be a metric space.

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

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

Then $M$ is complete.

Proof

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $S$.

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

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

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

Let $m > M$.

Let $n_1, n_2 > N$.

We have:

\(\ds \map d {y_{n_1, m}, y_{n_2, m} }\)

\(\le\)

\(\ds \map d {y_{n_1, m}, x_{n_1} } + \map d {x_{n_1}, x_{n_2} } + \map d {x_{n_2}, y_{n_2, m} }\)

two applications of the Triangle Inequality

\(\ds \)

\(<\)

\(\ds \epsilon\)

Therefore $\sequence {y_{m, n} }_{n \mathop \in \N}$ is Cauchy in $A$ for $m > M$.

So $\sequence {y_{m, n} }_{n \mathop \in \N}$ converges to some limit $y_n \in S$.

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.