# Is the Discrete Topology on $X$ the Only One Containing All Infinite Subsets of $X$?

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? – Gerry Myerson Nov 9 '12 at 5:02

## 2 Answers

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. – Cameron Buie Nov 9 '12 at 5:36
I think it is safe to assume the question takes the Axiom of Choice for granted. – madprob Nov 9 '12 at 5:44
And we rely on choice principles all the time, and what harm does it do? – Gerry Myerson Nov 9 '12 at 5:44
@madprob: I suspect you're right. – Cameron Buie Nov 9 '12 at 5:45

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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 9 '12 at 5:45
show 6 more comments

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. – M.Sina Nov 9 '12 at 7:42
You're welcome. Hopefully I didn't just confuse things for you. – Cameron Buie Nov 9 '12 at 18:44
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