Indiscrete Space is Pseudometrizable

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Theorem

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Then $T$ is pseudometrizable.


Proof

Let $d: S \times S \to \R$ be the mapping defined as:

$\forall x, y \in S: d \left({x, y}\right) = 0$

Then clearly $d$ is a pseudometric.


Let $\left({S, \tau_{\left({S, d}\right)}}\right)$ be the topological space induced by $d$.

Since $\left({S, \tau_{\left({S, d}\right)}}\right)$ is a topological space, by Empty Set is Element of Topology we have that $\varnothing$ is open.


Clearly, for any $\epsilon$ we have:

$B_\epsilon \left({a}\right) := \left\{{x \in S: d \left({x, a}\right) < \epsilon}\right\} = S$

where $B_\epsilon \left({a}\right)$ is the open $\epsilon$-ball of $a$.


Let $U\subset S$ be a non-empty open set.

By Open Sets in Pseudometric Space we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $B_\epsilon \left({x}\right)\subset U$.

However, $S\subset B_\epsilon \left({x}\right)$.

Thus $U = S$.

Clearly $\varnothing$ and $S$ are the only open sets of this resulting pseudometric space.

$\blacksquare$


Sources