Can spaces where all singletons are closed and all singletons are open be homeomorphic?

Question

Suppose $(X, \mathfrak{T})$ is a space where all singletons are closed, and $(Y, \mathfrak{J})$ is a space where all singletons are open.

Can these two spaces be homeomorphic? My thought is that they cannot be, but I am having a difficulty coming up with a proof.

A realistic example maybe $\mathbb{R}_{usual}$ and $\mathbb{R}_{discrete}$. But the thing is $\mathbb{R}_{discrete}$ all singletons are closed as well...

Answer 1

If all singletons are open, then the space is discrete, because an arbitrary union of open sets is an open set, and any set is a union of singletons. So the second space has the discrete topology. A discrete space is only homeomorphic to another discrete space whose underlying set has the same cardinality. So there are no "nontrivial" examples.

Answer 2

The preimage of a open set is still open. And the preimage of a singleton under a bijection is still a singleton. So the singletons is open in both spaces. A space with all singletons open can only have discrete topology.