# Can a finite set have a topology with an infinite number of open sets?

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?

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 $$\tau.$$
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.
• 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. Jul 7, 2020 at 20:50