Is there a countably infinite closed interval in the lattice of topologies?

Question

Is there an interval of the form $[\sigma,\tau]$ in the lattice of topologies on some set $X$ such that $|[\sigma,\tau]| = \aleph_0$? In other words, do there exist two topologies $\sigma$ and $\tau$ on $X$ such that there are a countably infinite number of topologies on $X$ that are finer than $\sigma$ and coarser than $\tau$?

I can find intervals like this of size $n$ for every finite $n > 0$, and intervals of size $\mathfrak{c}$. But I don't see how to obtain closed intervals of any intermediate size. I have a proof (well, an idea for a proof) that if $|[\sigma,\tau]| = \aleph_0$, then $\sigma$ cannot be a Hausdorff topology. Other than this, though, I can't seem to say much of anything about this question.

Answer 1

The special case where $\sigma=\{\emptyset,X\}$ is the trivial topology is easy to resolve. In this case, if $\tau$ is finite, then the interval $[\sigma,\tau]$ is finite. If $\tau$ is infinite, then there must exist an $\omega$-chain $T_0\subsetneq T_1\subsetneq T_2\subsetneq\cdots$ of nonempty, proper open sets in $\tau$ or a dual $\omega$-chain of nonempty, proper open sets in $\tau$. Assuming the former, for any nonempty subset $I\subseteq \omega$ let ${\mathcal B}_I=\{T_i\;|\;i\in I\}$ be a basis for a subtopology of $\tau$. Since $I\neq J$ implies that ${\mathcal B}_I$ and ${\mathcal B}_J$ generate different subtopologies of $\tau$, we get at least $\mathfrak c$ topologies in the interval $[\sigma,\tau]$.