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Let $(X,\tau)$ be a topological space. I need to show that if every subset is closed then it is a discrete space.

For finite $X$, let $S$ be a subset of $X$. Since $S$ is closed, $X \setminus S \in \tau$. But $X \setminus S \subseteq X$. Therefore, $X \setminus S$ is closed and $S$ is open. Since the choice of $S$ was arbitrary, $(X,\tau)$ is a discrete space.

My doubt is whether the proof holds for $X$ being infinite too.

Parth
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    Your proof is okay but contains a redundancy. The obervation that $S$ is closed and consequently that its complement is open is not really used in your proof, and can be left out. Yes, the proof holds also if $X$ is infinite. – drhab May 06 '17 at 12:40

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Your proof looks good to me. Actually you did not use the fact that $X$ is finite. The same logic applies to any $X$.

$\forall s \subset X$, $X \setminus s \subset X$, and thus $X \setminus s $ is closed. Thus $s \in \tau$. - we are not involving anything about the finiteness of $X$.

Jay Zha
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  • Does the same logic apply to both finite and infinite subsets of $X$? – Parth May 06 '17 at 12:28
  • @Parth yes, it does. On your proof, which step did you think not work for infinite set? - Are you concerning "arbitrary choice of subset s from an infinite set X"? Notice we are not involving any additional axiom here - we just say "given any set". – Jay Zha May 06 '17 at 12:29
  • https://math.stackexchange.com/questions/677459/let-x-be-an-infinite-set-with-a-topology-t-such-that-every-infinite-subset?rq=1 here one of the comments states that $X \setminus S$ is closed only when it is finite. – Parth May 06 '17 at 12:34
  • @Parth Ah, get it. So for the link, it says every infinite subset of X is closed. That's is different, meaning finite subset is not guaranteed to be closed. However, you do not have the issue here, every subset is closed by your question. – Jay Zha May 06 '17 at 12:35
  • So shouldn't the comment state that $X \setminus S$ is closed only when it is infinite as the question guarantees that all infinite subsets are closed? – Parth May 06 '17 at 12:38
  • @Parth Yes, you are right, meant for that. Meaning it is possible that a subset of X or its compliment is not necessarily closed, because you might end up with a finite one, and there is no guarantee that it is closed. – Jay Zha May 06 '17 at 12:41
  • Maybe he meant the classical exercise: if all infinite subsets are closed then $X$ is discrete? – Henno Brandsma May 06 '17 at 17:51
  • @HennoBrandsma yes, and he posted the link in above comment, math.SE already has a post, and the answer is clear there, so I'm not gonna repeat here. – Jay Zha May 06 '17 at 17:53
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Let $x$ be a point of $X$, $X-\{x\}$ is closed so its complementary subset $\{x\}$ is open. Thus every subset is open since it is the union of its elements.

Tsemo Aristide
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Why wouldn't the proof work for infinite sets/spaces? Take any subset $\;A\subset X\;$, then $\;X\setminus A\;$ is closed and thus $\;A\;$ is open, and thus any subset of $\;X\;$ is open and we have the discrete topology...

DonAntonio
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