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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

Björn Friedrich
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ZZS14
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3 Answers3

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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,

Patrick Da Silva
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  • But how do we know that ${x} \in \tau$? – zermelovac Nov 11 '15 at 20:43
  • @zermelovac : I have proven that. I have shown that $X \backslash {x}$ is infinite, hence closed, so that ${x}$ is open. – Patrick Da Silva Nov 11 '15 at 21:11
  • So if ${x }$ is open, then ${x }\in T$.

    I know that:

    $(X,\tau)$ discrete topology $\Leftrightarrow$ $(\forall x \in X); {x}\in \tau$.

     – zermelovac Nov 11 '15 at 21:18
  • Can you explain your conclusion that $T$ is all subsets of $X$. Maybe then, I will understand. – zermelovac Nov 11 '15 at 21:23
  • @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. – Patrick Da Silva Nov 11 '15 at 21:56
  • I have understood that. The question a have is:

    If all the subsets of $X$ are open, then $\tau$ is the discrete topology?

     – zermelovac Nov 11 '15 at 22:02
  • Is it possible to use AofC since X is infinite –  Dec 13 '20 at 17:26
  • @Eudoxus Why would you want to use the axiom of choice here? I didn't need it. – Patrick Da Silva Dec 14 '20 at 10:16
  • X is infinite set,and has infinite subset –  Dec 14 '20 at 13:20
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    @Eudoxus: It's not because you work with infinite sets that you need the axiom of choice. The axiom of choice says that if you have a collection of sets (indexed by a set), you can pick one element in each set of your collection. In other words, if {X_i}{i in I} is a set, then the product prod{i in I} X_i is not empty, so you get an I-tuple of elements of each X_i. But here we don't need such a thing. It is true by the axiom of extensionality that the two sets above are equal (they contain the same elements) and since the union on the right is a union of open sets, S is open. – Patrick Da Silva Dec 16 '20 at 22:23
  • Did you get the email I sent you about learning different math topics @PatrickDaSilva –  Dec 17 '20 at 15:37
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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 $\boxed{T=\mathcal P(X)}$

xavierm02
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  • You already know that $T$ is a topology, thus it is clear that $\varnothing \in T$. – Patrick Da Silva Feb 15 '14 at 18:00
  • @xaviermo2 But how do we know that ${x} \in T$? – zermelovac Nov 11 '15 at 20:43
  • @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 ${x}\in T$. – xavierm02 Nov 12 '15 at 08:14
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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.

Sitos
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  • Please don't go back to 5-year-old questions with multiple good answers and provide another answer that doesn't add anything new. – Kevin Carlson May 20 '19 at 21:17