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Prove or find counterexamples.

Let $X$ be an infinite set and $T$ be a topology on $X$. If $T$ contains every infinite subset of $X$, then $T$ is the discrete topology.

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    Can you find two infinite sets whose intersection is finite?2012-11-09
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    @M.Sina: see here http://math.stackexchange.com/questions/60269/question-regarding-subsets-of-infinite-sets2013-06-06
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    @MaisamHedyelloo: Mamnoon Maisam jan :).2013-06-06
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    @M.Sina:your welcome M.sina2013-06-06

2 Answers 2

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Suppose $T$ is a topology containing all the infinite subsets of $X$. I claim every finite subset also belongs to $T$, and so $T$ is the discrete topology.

To see this, let $A$ be any finite subset of $X$. Since $X$ is infinite, $X \setminus A$ is infinite. Partition $X \setminus A$ into two disjoint infinite subsets $Y_1$ and $Y_2$ (this can always be done if the Axiom of Choice is assumed).

Now, $Y_1 \cup A$ and $Y_2 \cup A$ are both infinite sets, so they belong to $T$. Moreover, their intersection is precisely $A$. Since topologies are closed under finite intersection, it must be the case that $A$ belongs to $T$. Since $A$ was an arbitrary finite set, the claim follows.

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    Ever heard of amorphous sets? We can't necessarily partition infinite sets into two disjoint infinite subsets without reliance on some choice principle.2012-11-09
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    I think it is safe to assume the question takes the Axiom of Choice for granted.2012-11-09
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    And we rely on choice principles all the time, and what harm does it do?2012-11-09
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    @madprob: I suspect you're right.2012-11-09
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    @GerryMyerson: Certainly, no harm is done by using them. On the other hand, no harm is done by noting when they're used, just in case they aren't meant to be for some reason.2012-11-09
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    @Austin mohr: Please get more tips, why Partition X∖A into two disjoint infinite subsets Y1 and Y2 ?2012-11-09
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    @MohammaDSina My goal was to find two open sets whose intersection would be precisely the finite set $A$. These sets would need to be infinite (to ensure they belong to $T$) and disjoint except for $A$ (to ensure the intersection is precisely $A$). The sets $Y_1 \cup A$ and $Y_2 \cup A$ are simply the two sets I came up with (though, I don't think there is much else one can do here).2012-11-09
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    @AustinMohr Thanks for your detailed response.2012-11-09
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    @Cameron: I shouldn’t worry about it, if I were you: at the undergraduate level it is essentially never the case that anyone is supposed to worry about choice. Introducing that issue is more likely to result in confusion than in enlightenment.2012-11-09
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    @Brian: I had been thinking in terms of accessible and complete answers for other users who might be better versed in questions of choice principles. I take your point, though, about confusing the issue. I'll try to bear that in mind in the future.2012-11-09
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    @Cameron: I freely acknowledge that my first concern is almost always for the OP, with future readers a fairly distant second. And the few exceptions almost always involve questions on comparatively advanced topics.2012-11-09
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Let's suppose that $X$ is an amorphous set, in the cofinite topology. By the amorphous nature of $X$, every infinite subset of $X$ has a finite complement, so every infinite subset of $X$ is open. Thus, the open subsets of $X$ are precisely the empty set, $X$, and the infinite proper subsets of $X$. However, this is not discrete, as (for example) no singleton subset of $X$ is open.

Now, if we have enough Choice so that there aren't any amorphous sets, then Austin's approach is the way to go for a proof. Otherwise, the above serves as a counterexample.

Remark: I don't intend this to be a "competing" answer with Austin's. I intended merely to elucidate why I brought up the Axiom of choice and why he then felt compelled to make mention of it in his answer. He answered before I did, it's a good answer, and you didn't pre-specify how much (if any) Choice you're using. If you like mine, feel free to upvote, but if you're debating which of our answers to accept, go with his.

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    Thanks for your comments.2012-11-09
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    You're welcome. Hopefully I didn't just confuse things for you.2012-11-09
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    @downvoter: Is there a problem with this answer that I can address?2014-12-03