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Let $X$ be an infinite set with a topology $T$, such that every infinite subset of $X$ is closed. Prove that $T$ is the discrete topology.

I have somewhat of an answer but I don't think it's enough to prove it, especially with respect to the subsets being infinite.

Let $S$ be contained in $X$, then $X \setminus S$ is also contained in $X$. Therefore we can say that $X \setminus S$ is closed, therefore $S$ is open for any $S$ contained in $X$. Hence $T$ is the discrete topology.

Thanks

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  • $\begingroup$ "Therefore we can say that $X \setminus S$ is closed" why? $\endgroup$ – Najib Idrissi Feb 15 '14 at 15:45
  • $\begingroup$ You can say that $X\setminus S$ is closed only in the case where $X\setminus S$ is finite, which doesn't have to happen. $\endgroup$ – kneidell Feb 15 '14 at 15:46
  • $\begingroup$ @nik : This is not the same problem, the question you linked says that every infinite subset of $X$ is -open-, not closed. $\endgroup$ – Patrick Da Silva Feb 15 '14 at 15:49
  • $\begingroup$ @PatrickDaSilva: You're right, my apologies. $\endgroup$ – Najib Idrissi Feb 15 '14 at 15:49
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If $X$ is infinite, let $x \in X$ ; it follows that since $\{x\}$ is finite, $X \backslash \{x\}$ is infinite, hence closed. Therefore, for any subset $S \subseteq X$, $$ S = \bigcup_{x \in S} \{x\} $$ is a union of open sets, thus open. This means $T$ is all subsets of $X$. You have the discrete topology.

Hope that helps,

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  • $\begingroup$ But how do we know that $\{x\} \in \tau$? $\endgroup$ – zermelovac Nov 11 '15 at 20:43
  • $\begingroup$ @zermelovac : I have proven that. I have shown that $X \backslash \{x\}$ is infinite, hence closed, so that $\{x\}$ is open. $\endgroup$ – Patrick Da Silva Nov 11 '15 at 21:11
  • $\begingroup$ So if $\{x \}$ is open, then $\{x \}\in T$. I know that: $(X,\tau)$ discrete topology $\Leftrightarrow$ $(\forall x \in X)\; \{x\}\in \tau$. $\endgroup$ – zermelovac Nov 11 '15 at 21:18
  • $\begingroup$ Can you explain your conclusion that $T$ is all subsets of $X$. Maybe then, I will understand. $\endgroup$ – zermelovac Nov 11 '15 at 21:23
  • $\begingroup$ @zermelovac : Any subset is the union of its singleton subsets and arbitrary unions of open subsets are open. Therefore if singletons are open, any subset of $X$ is open. $\endgroup$ – Patrick Da Silva Nov 11 '15 at 21:56
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$X$ is infinite so it is closed so $\emptyset = X^c$ is open. $\boxed{\emptyset \in T}$

$X\setminus\{x\}$ is infinite so $\{x\}$ is open. $\boxed{\forall x \in X,\{x\}\in T}$

And for $A \subseteq X$ where $A\not= \emptyset$, $A=\bigcup\limits_{x\in A}\{x\}$. So $\boxed{\mathcal P(X)\setminus\{\emptyset\}\subseteq T}$

So $\boxed{T=\mathcal P(X)}$

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  • $\begingroup$ You already know that $T$ is a topology, thus it is clear that $\varnothing \in T$. $\endgroup$ – Patrick Da Silva Feb 15 '14 at 18:00
  • $\begingroup$ @xaviermo2 But how do we know that $\{x\} \in T$? $\endgroup$ – zermelovac Nov 11 '15 at 20:43
  • $\begingroup$ @zermelovac $T$ is the set of open sets. Since $\{x\}$ is finite and $X$ is infinite, $X\setminus \{x\}$ is infinite and is therefore closed. And saying that $X\setminus \{x\}$ is closed is the same thing as saying that $\{x\}$ is open. So $\{x\}\in T$. $\endgroup$ – xavierm02 Nov 12 '15 at 8:14
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Every infinite subset of X closed means every finite subset of X is open. In particular, the singleton sets {x} for all x in X are open. This is sufficient to conclude that T is discrete!

Hope it helps.

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  • $\begingroup$ Please don't go back to 5-year-old questions with multiple good answers and provide another answer that doesn't add anything new. $\endgroup$ – Kevin Carlson May 20 at 21:17

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