When is $\mathscr{T}_f=\mathscr{T}_c$ exactly?

Question

BACKGROUND

For any set $X$ and $\mathscr{T}$ a topology on $X$, $\mathscr{T}$ is finer than the indiscrete (or trivial) topology $\mathscr{T}_i$ on $X$. The discrete topology $\mathscr{T}_d$ on $X$ is finer than $\mathscr{T}$. That is, we have $$\mathscr{T}_i \subseteq \mathscr{T} \subseteq \mathscr{T}_d.$$ We have the equality $$\mathscr{T}_i = \mathscr{T} = \mathscr{T}_d$$ when the set $X = \emptyset$ or when $X$ is a singleton.

Similarly, the cocountable topology $\mathscr{T}_c$ is finer than the cofinite topology $\mathscr{T}_f$, or $$\mathscr{T}_f \subseteq \mathscr{T}_c,$$ because every finite set is a countable set.

PROBLEM

When is $\mathscr{T}_f=\mathscr{T}_c$ exactly?

MY ATTEMPT

Equality occurs in $$\mathscr{T}_f \subseteq \mathscr{T}_c$$ when $X = \emptyset$ or when $X$ is finite, again because every finite set is a countable set, but not all countable sets are finite.

QUESTION

Is my intuition (as in my answer) correct?

Answer 1

You are correct!

The cocountable and cofinite topologies on $X$ are equal iff $X$ is finite.

Proof: If $X$ is infinite, then there is a countably infinite subset $D$ of $X$. (This uses the axiom of choice.) Then $D^c$ is cocountable, but not cofinite. Conversely, if $X$ is finite, then every subset of $X$ is both cofinite and cocountable. This means that the cofinite, cocountable, and discrete topologies coincide.