Discrete Space is Complete Metric Space

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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 get that $(S,d)$ is a metric space whose induced topology is $\tau$.

Consider now a Cauchy sequence $\left({x_n}\right)_{n \in \N}$.

By the definition of Cauchy sequence, $\forall \varepsilon > 0: \exists N \in \N$ such that $\forall n, m > N: d (x_n, x_m) < \varepsilon$.

Take $\varepsilon = \dfrac 1 2$.

Then $\exists N \in \N$ such that $\forall n, m > N: d (x_n, x_m) < \varepsilon = \dfrac 1 2$.

From the definition of discrete metric, $\exists N \in \N$ such that $\forall n, m > N: d (x_n, x_m) = 0 \implies x_n = x_m$.

Then, the sequence $\left({x_n}\right)_{n \in \N}$ is constant from some $N\in \N$.

Since every eventually constant sequence converges, $x_n \to x \in S$.

Thus, every Cauchy sequence converges.

Hence the result, by definition of complete metric space.

$\blacksquare$


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