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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.

Austin Mohr
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M.Sina
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2 Answers2

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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.

Austin Mohr
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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. – Cameron Buie Nov 09 '12 at 05:36
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    I think it is safe to assume the question takes the Axiom of Choice for granted. – madprob Nov 09 '12 at 05:44
  • And we rely on choice principles all the time, and what harm does it do? – Gerry Myerson Nov 09 '12 at 05:44
  • @madprob: I suspect you're right. – Cameron Buie Nov 09 '12 at 05:45
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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. – Cameron Buie Nov 09 '12 at 05:45
  • @Austin mohr: Please get more tips, why Partition X∖A into two disjoint infinite subsets Y1 and Y2 ? – M.Sina Nov 09 '12 at 07:40
  • @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). – Austin Mohr Nov 09 '12 at 07:44
  • @AustinMohr Thanks for your detailed response. – M.Sina Nov 09 '12 at 08:07
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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. – Brian M. Scott Nov 09 '12 at 15:33
  • @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. – Cameron Buie Nov 09 '12 at 18:43
  • @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. – Brian M. Scott Nov 09 '12 at 18:51
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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.

Cameron Buie
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