Discrete Space is Complete Metric Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space where $\tau$ is the discrete topology on $S$.

Then $T$ is a complete metric space.

Proof
Let $d: S \times S \to \R$ be the discrete metric.

From Discrete Metric induces Discrete Topology we have that $\left({S, d}\right)$ is a metric space whose induced topology is $\tau$.

Consider now a Cauchy sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$.

By the definition of Cauchy sequence:
 * $\forall \epsilon > 0: \exists N \in \N$ such that $\forall n, m > N: d \left({x_n, x_m}\right) < \epsilon$

Take $\epsilon = \dfrac 1 2$.

Then $\exists N \in \N: \forall n, m > N: d \left({x_n, x_m}\right) < \epsilon = \dfrac 1 2$

From the definition of discrete metric:
 * $\exists N \in \N: \forall n, m > N: d \left({x_n, x_m}\right) = 0 \implies x_n = x_m$

Thus the sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ is constant from some $N \in \N$.

From Eventually Constant Sequence is Convergent:
 * $x_n \to x \in S$

Thus every Cauchy sequence in $S$ converges.

The result follows by definition of complete metric space.