Can $(X, τ)$ be a nontrivial topological space such that $(τ,X)$ is also a topological space and a metric space under the same metric?

Question

Can $(X, τ)$ be a nontrivial topological space such that $(τ,X)$ is also a topological space and a metric space under the same metric?

I'm looking at the Collatz graph as a surjection from a subset of $\omega^\omega$ onto itself and it appears to have this property.

FWIW my thinking about the answer to this is that it would require $τ$ to be a family of sets drawn from the powerset of $X$ and vice versa, and therefore requires $τ=X$. Would this be correct?

EDIT: This only seems possible with non-well-founded forms of set theory as it requires breaking Russel's paradox. Is this correct, and e.g. can $X=\{X\}$ have this topology?

Answer 1

As you already noted, this cannot work in any well-founded set theory; in particular, it won't work in ZF (with or without choice).

However in other set theories, it is indeed possible. For example, when replacing the axiom of foundation with Aczel's anti-foundation axiom, the set you've brought up in the comments, $$X = \{\emptyset,X\}$$ is indeed an example (actually, the simplest one). It gives the trivial topology over itself.

With anti-foundation, $X$ is a set because it corresponds to a two-node graph where one node has a loop and an arrow to the other node. The node with the loop is the root.

Answer 2

Reposting my comment as an answer.

By definition of topology you have $X\in\tau$. If $(\tau,X)$ were a topological space too you would have $\tau\in X$ as well, but it's impossible to have such loops of membership as a consequence of the axiom of foundation.