I am trying to show that $[0,1]$ is not compact when $\mathbb{R}$ is equipped with the discrete metric.

Consider $\mathbb{R}$ equipped with the discrete metric. Since every singleton set in $\mathbb{R}$ is both open and closed, under the discrete metric, every subset of $\mathbb{R}$ is both open and closed. Thus $[0,1]$ can be expressed as the union of arbitrarily many singleton sets (the union of arbitrarily many open sets is itself open). The collection of such sets constitutes an open cover of $[0,1]$ (call this cover $G$). Yet, if we remove at most one of the singleton sets from $G$, $G$ no longer covers $[0,1]$. Thus, $G$ is a cover, of $[0,1]$, that does not admit a finite sub cover.

Thank you.

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    $\begingroup$ Looks good to me! $\endgroup$ Aug 30, 2021 at 16:28
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    $\begingroup$ I don't think you actually need the second sentence. Your argument only relies on the fact that singleton sets are open, which is true by definition in the discrete topology. You don't really need to talk about other sets being open or closed. (That doesn't make the proof wrong, of course — it just could be tightened up.) $\endgroup$ Aug 30, 2021 at 17:14

1 Answer 1


Your solution looks fine. In fact, you can try proving the following statement in general:

Let $X$ be a set with the discrete metric. Then, $X$ is compact iff $X$ is finite.

Note that finite topological spaces are always compact. The above shows that if we assume the topology to be discrete, then it's a two-way implication.


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