Is a countable, totally disconnected Hausdorff space necessarily totally separated? How about zero-dimensional?

I'm starting to feel a little bad about using this website as my own personal counterexample generator, but here I go again...

Terminology:

Let's call a space zero-dimensional if it is $T_0$ and admits a basis of clopen sets. By a standard embedding argument, a space is zero-dimensional if and only if it is homeomorphic to a subspace of some (possibly uncountable) power of the two-point discrete space $\{0,1\}$. In particular, zero-dimensional implies Hausdorff, or even completely regular.

Let's call a space $X$ totally separated if, given distinct points $x,y \in X$, there exists a separation $U,V$ of $X$ (ie. $U,V$ partition $X$ and are open) such that $x \in U$ and $y \in V$. In particular, a totally separated space is Hausdorff. Clearly zero-dimensional implies totally separated.

Finally, let's call a space totally disconnected if all its connected components are singletons. Clearly totally separated implies totally disconnected (conversely, totally disconnected need not even imply Hausdorff).

My question:

Let $X$ be a countable, totally disconnected Hausdorff space. Can $X$ fail to be totally separated? If yes, can $X$ fail to be zero-dimensional?

Some discussion:

If we replace "countable" with "compact, the answer to both questions is "no". For a compact Hausdorff space, the components and quasicomponents coincide, so $X$ is totally separated. Then, by applying basic compactness arguments, we can prove even that for all $A,B \subset X$, disjoint closed sets, there is a separation $U,V$ of $X$ with $A \subset U, B \subset V$. In particular, $X$ is zero-dimensional. The hypothesis of compactness cannot be dropped though. For example Cantor's leaky tent is a (noncompact, noncountable) subspace of the Euclidean plane which can be shown, with some effort, to be totally disconnected - but not totally separated. Since the hypothesis of compactness cannot be dropped, I wondered whether it could be replaced with something else. In particular, I wondered whether countable would do.

Added: Here's another counterexample. The main idea is the same as in Brian's example, but I thought this space seemed somehow more concrete.

As a set, let $X := \mathbb{Q} \cup \{p_0,p_1\}$ where $p_0,p_1$ are two distinct points not in $\mathbb{Q}$. For $n=0,1,2,\ldots$, let $I_n := (n,n+1) \cap \mathbb{Q}$. We put $U \subset X$ open if and only if the following are satisfied:

1. $U \cap \mathbb{Q}$ is open in the standard topology on $\mathbb{Q}$.
2. If $p_0 \in U$, then $U$ contains all but finitely many of $I_0,I_2,I_4,\ldots$
3. If $p_1 \in U$, then $U$ contains all but finitely many of $I_1,I_3,I_5,\ldots$

It is easy to see this topology is Hausdorff. To see it is totally disconnected, suppose that $C \subset X$ is connected with more than one point. Intervals with irrational endpoints are still clopen, and these can be used to separate a fixed rational number from any other point in $X$. It follows that $C$ contains no rationals. Thus $C = \{p_0,q_0\}$, but this space is discrete ($X$ is Hausdorff), so no such $C$ exists. However, $X$ is not totally separated. Neighbourhoods of $p_0$ and $q_0$ cannot have disjoint closures.

Solution 1:

Let $X=(\omega\times\mathbb{Z})\cup\{p^-,p^+\}$, where $p^-$ and $p^+$ are distinct points not in $\omega\times\mathbb{Z}$. Let $Z_0=\mathbb{Z}\setminus\{0\}$. Points of $\omega\times Z_0$ are isolated. For each $n\in\omega$ and finite $F\subseteq Z_0$ let $$B(n,F)=\{n\}\times(Z_0\setminus F)\;,$$ and take $\{B(n,F):F\subseteq Z_0\text{ is finite}\}$ as a local base at $\langle n,0\rangle$. For $n\in\omega$ let $$B^+(n)=\{p^+\}\cup\{\langle i,k\rangle\in\omega\times Z_0:i>n\land k>0\}$$ and $$B^-(n)=\{p^-\}\cup\{\langle i,k\rangle\in\omega\times Z_0:i>n\land k<0\},$$ and take $\{B^+(n):n\in\omega\}$ and $\{B^-(n):n\in\omega\}$ as local bases at $p^+$ and $p^-$, respectively.

It’s easy to check that $X$ is Hausdorff and totally disconnected. However, $p^-$ and $p^+$ do not have open nbhds with disjoint closures: for any $n,m\in\omega$, $$\operatorname{cl}B^-(n)\cap\operatorname{cl}B^+(m)\supseteq\big\{\langle k,0\rangle:k>\max\{n,m\}\big\}\;.$$ It follows immediately that $X$ is neither totally separated nor zero-dimensional.

Solution 2:

In Steen & Seebach, Counterexamples in Topology, p. 99, they prove that the Arens square is also an example of a countable, totally disconnected Hausdorff space that is neither totally separated nor zero dimensional. I can post some details if nobody has the reference on hand.