I came across the following exercise: suppose we have a topology on an infinite set $X$ which contains all infinite subsets of $X$. Prove that the topology is discrete.

Here's a way to approach this: pick two disjoint infinite subsets $A$ and $B$ of $X$. For any $x\in X$, $A\cup\left\{x\right\}$ and $B\cup\left\{x\right\}$ are open and so is their intersection, which is just $\left\{x\right\}$. Hence all singletons are open and the topology is discrete.
This proof is essentially the same as the one here.

However, I don't quite like this argument, mainly because of the step where we say "pick two disjoint infinite subsets of $X$". This isn't really a problem in ZFC since it's possible to prove that every infinite set has an infinite countable subset and after this is established, picking $A$ and $B$ is simple. But what if I'm working without the axiom of choice?

So, my question is: can it be proved without AC that every infinite set contains two disjoint infinite subsets or is it perhaps possible to avoid this altogether by proving the topology is discrete in another way (not reliant on AC)?

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    $\begingroup$ If I recall correctly (this isn't an answer because I'm not sure) it is consistent with ZF that there is an infinite set such that every subset is finite or cofinite. If such a set exists, then the set of infinite subsets of $X$ (+ $\emptyset$) is a topology (stable under finite unions because cofinite sets are), and it's not discrete $\endgroup$ Commented Jun 20, 2019 at 17:28

1 Answer 1


No, it is (relatively) consistent with ZF that there exists an infinite set that does not have two disjoint infinite subsets. Such as set is called amorphous.

If $X$ is an amorphous set, then the topology you describe is the cofinite topology, which is not discrete.

So you need some form of choice to prove your goal.

  • $\begingroup$ Interesting.... $\endgroup$
    – Randall
    Commented Jun 20, 2019 at 17:29
  • $\begingroup$ This is surprising! I guess with or without AC, strange things happen $\endgroup$
    – J_P
    Commented Jun 20, 2019 at 17:33
  • $\begingroup$ @J_P without, more strange things, I think. $\endgroup$ Commented Jun 20, 2019 at 22:30
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    $\begingroup$ @DanielWainfleet So if I understand correctly, there exists a model of ZF (a model being a particular realization of ZF) in which, say, amorphous sets exist, but there is no way to prove their existence just from ZF axioms alone. $\endgroup$
    – J_P
    Commented Jun 21, 2019 at 14:25
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    $\begingroup$ @J_P: Correct -- you can't prove the existence of such sets from ZF, because such a proof would still be valid in ZFC (adding more axioms can never rob an existing proof of validity) and create a contradiction with AC proving that amorphous sets don't exist. $\endgroup$ Commented Jun 21, 2019 at 14:30

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