Totally disconnected topologies on countable set.

Question

Are there totally disconnected topologies $\tau$ on a countable set $X$ such that $(X,\tau)$ is not homeomorphic to one of the following?

• $\mathbb{N}$ with the discrete topology;
• one-point compactification of $\mathbb{N}$ with the discrete topology;
• $\mathbb{Q}$ with the Euclidean topology;
• $\mathbb{Z}$ with the $p$-adic topology for some prime $p$

?

Answer 1

If $\mathcal{F}$ is a free ultrafilter on $\omega$, we can define a topology $\mathcal{T}(\mathcal{F})$ on $\omega \cup \{p\}$, where all points of $\omega$ are still isolated and neighbourhoods of $p$ are $\{\{p\} \cup A: A \in \mathcal{F}\}$. This defines a scattered topology on a countable set, which is hereditarily normal, but not metrisable.

The spaces are homeomorphic (for different ultrafilters) iff they ultrafilters are equivalent (there is a permutation of $\omega$ that transforms one in the other); there are $2^{2^{\omega}}$ many non-equivalent ultrafilters on $\omega$, so that many non-homeomorphic such ultrafilter spaces.

Answer 2

The obvious two point compactification of $\Bbb Z$ does the trick.

Answer 3

Given any countable ordinal $\alpha$, the order topology is a totally disconnected topology (it is homeomorphic to a subspace of the rationals).

If two ordinals are homeomorphic then they differ by finitely many isolated points. So different limit ordinals will give different spaces. There are uncountably many countable limit ordinals, which provides us with plenty of examples.

You get even more. Every successor ordinal is a compact space, so you get uncountably many compact examples that way too.