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)?
Best Answer
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.