Can a finite set have a topology with an infinite number of open sets? ..(1)

The question originated when my professor gave us as an example that if $X$ is finite or $\tau$ is finite, $(X, \tau)$ is compact

And that that was so, even if, in the case of finite $X$ , $\tau$ had an infinite number of open sets... Now if the topology has a finite number of open sets it is clear we can always extract a finite subcover, which is the initial cover itself, but what about the infinite case, why is it true, provided (1) is possible?


2 Answers 2


No. Consider that any topology $\tau$ on a set $X$ will be a subset of the powerset, that is $\tau \subseteq P(X).$ Since $X$ is finite, then $P(X)$ is finite, and consequently so is $\tau.$


A finite set can only have finitely many subsets. Therefore every topology on a finite set is finite itself.

What may confuse you, however, is that the proof that every finite space is compact does not go through "the power set is finite, therefore we are done", but rather by saying that if $\{U_i\mid i\in I\}$ is an open cover, then for every $x\in X$ we can choose some $i_x$ such that $x\in U_{i_x}$, and therefore $\{U_{i_x}\mid x\in X\}$ is a finite subcover.

This is a "more correct" proof, because it actually shows that every finite set is compact in every topology. Even if the space itself is infinite, every finite subset is compact.

  • $\begingroup$ If the set is finite, can't I take as finite subcover the cover itself, that it's already finite? or should the subcover be always a proper set of the cover? $\endgroup$ Jul 7, 2020 at 20:49
  • $\begingroup$ Well, $\{(-n,n)\mid n\in\Bbb N\}$ is an open cover of $\{0\}$ in $\Bbb R$ (with the standard topology), of course any of the intervals can be taken as a finite subcover. But that is an infinite cover of a finite set. $\endgroup$
    – Asaf Karagila
    Jul 7, 2020 at 20:50
  • $\begingroup$ I guess that is what my professor meant , when he said a finite set and infinite open sets on the topology. . If I take as cover of {0} a single set, say (-2,2) , does it make sense to say that (-2,2) itself is the finite subcover? or to talk about subcover we need a proper subset of the cover, that in this case doesn't exist, since we have only one set $\endgroup$ Jul 7, 2020 at 20:59
  • $\begingroup$ Subcover is a subset. Subsets are not necessarily proper subsets. $\endgroup$
    – Asaf Karagila
    Jul 7, 2020 at 20:59

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