Indiscrete Topology is not Metrizable

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Theorem

Let $S$ be a set with more than one element.

The indiscrete topology on $S$ is not metrizable.


Proof

In order to be metrizable, there needs to be a metric $d$ on $S$, so that $\left({S, d}\right)$ is a metric space.

As $S$ has more than one element, $\exists x, y \in S: x \ne y$.

Then $\epsilon = d \left({x, y}\right) > 0$.

So the open $\epsilon$-ball $B_{\epsilon / 2} \left({x}\right)$ is $d$-open.

Hence $B_{\epsilon / 2} \left({x}\right)$ is in the topology which $d$ induces.

But $x \in B_{\epsilon / 2} \left({x}\right)$ while $y \notin B_{\epsilon / 2} \left({x}\right)$.

Thus $B_{\epsilon / 2} \left({x}\right) \ne \varnothing$ and $B_{\epsilon / 2} \left({x}\right) \ne S$.

So the topology induced by $d$ is not the indiscrete topology.

$\blacksquare$


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