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I want to prove the following: Let $X$ be an infinite set and $\tau$ a topology on $X$. If every infinite subset of $X$ is in $\tau$, then $\tau$ is the discrete topology on $X$.

Proof. Let $x\in X$. There exist two infinite subsets $A$ and $B$ of $X$ such that $\{x\}=A\cap B$. So every singleton is open in $X$. It follows that $\tau$ is the discrete topology on $X$.

Edit. Justifying the existence of $A$ and $B$. Assume that $C$ is a countable subset of $X$. Let $A$ be the set of odd-numbered points in $C$ and $B=(C-A)\cup\{x\}$ for some $x$ in $A$ (i.e. $B$ contains the even-numbered points together with $x$). $A\cap B=\{x\}$. This holds for every $x$ in $B$ as well (by symmetry) so such $A$ and $B$ always exist.

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marked as duplicate by Amzoti, Daniel Rust, Aang, azimut, O.L. Jul 15 '13 at 13:08

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what is the question? –  Thomas Rot Jul 15 '13 at 12:43
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The proof looks fine. I think you may need to justify your statement that such an $A$ and $B$ exist though. –  Daniel Rust Jul 15 '13 at 12:43
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@ThomasRot Is the proof correct? –  saadtaame Jul 15 '13 at 12:43
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as daniel suggests, the existence of $A$ and $B$ is the only conceivable missing detail. think hilbert's hotel –  citedcorpse Jul 15 '13 at 12:44

1 Answer 1

up vote 1 down vote accepted

Hints:

Choose an arbitrary $\,x\in X\;$ :

== There exists an infinite countable $\,Y\subset X\setminus\{x\}\,$ , say $\,Y=\{y_1,y_2,\ldots\}\;$

== Define $\,A:=\{y_i\in Y\;;\;i\;\;\text{is an odd natural number}\}\cup\{x\}\\B:=\{y_i\in Y\;;\;i\;\;\text{is an even natural number}\}\cup\{x\}$

== Deduce that any point in $\;X\;$ is open ...

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