# 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.

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I don't know nothing about topological spaces but I know that sometimes when you don't have much ideas to get counter examples for stuff it can be a pain. Sometimes people just have more things than you do in mind and it's a good thing to ask those things here, but if you want your answers to be more useful for you in the future, perhaps you should not only ask for a counter example but also how it was generated, so that you can do a better job in the future. =) – Patrick Da Silva Nov 20 '11 at 0:44
The reference book for counterexamples in topology is called, surprisingly, Counterexamples in Topology. It is by Steen & Seebach. – GEdgar Nov 20 '11 at 1:30
@GEdgar: There's a never ending "recall war" for my university library's copy of that book. I don't remember there being an example of the type I'm looking for, but I could easily be mistaken. Can somebody with a copy on hand confirm? – Mike F Nov 20 '11 at 3:00

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.

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This is fantastic! It's closely related to the construction in this paper but with all the extraneous baggage (ie. the filter of cofinite sets in place of the unnecesary nonprinciple ultrafilter) stripped away. With only one ideal point adjoined, your space would still be totally separated, but you managed to fix this by introducing an extra point. Really well done! – Mike F Nov 20 '11 at 7:45
I added another counterexample (based on yours) at the end of the problem statement, if you'd care to have a peek. – Mike F Nov 20 '11 at 10:25
@Mike: Yes, that works very nicely. – Brian M. Scott Nov 20 '11 at 10:28

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.

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That's ok, I can look it up on monday, thanks! – Mike F Nov 20 '11 at 7:33

Some vigorous googling turned up a paper containing an example of a countable, totally disconnected Hausdorff space which is not regular (hence not zero-dimensional). They use the example to demonstrate something rather more elaborate, so I have a feeling there is a lot of room to simplify the construction which I'll outline below.

As a set, we take $X$ to be the disjoint union of countably many copies of $\mathbb{N}$ (which I take to include $0$), one distinguished copy of $\mathbb{N}$, and an idealized point $p$. So, $$X = (\mathbb{N}_0 \cup \mathbb{N}_1 \cup \mathbb{N}_2 \ldots ) \cup \mathbb{N} \cup \{p\}.$$ Now to topologize $X$. For this, we will need to fix an nonprincipal ultrafilter $\mathscr{P} \subset 2^\mathbb{N}$. For each $i$, let $\mathscr{P}_i$ denote the corresponding copy of $\mathscr{P}$.

• Each point in each $\mathbb{N}_i$ is taken to be isolated.
• The neighbourhoods of $i \in \mathbb{N}$ are the sets that contain $\{i\} \cup P_i$ for some $P_i \in \mathscr{P}_i$.
• The neighbourhoods of $p$ are the sets that $\{p\} \cup \bigcup_{i \in P} \mathbb{N}_i$ for some $P \in \mathscr{P}$.

It is clear the open sets so defined are closed under unions. To show that the intersection of two neighbourhoods some $i \in \mathbb{N}$ or two neighbourhoods of $p$ is still a neighbourhood one just needs that filters are closed under finite intersections.

The topology we obtain is Hausdorff. For instance, we can separate $i \in \mathbb{N}$ from $x \in \mathbb{N}_i$ because $\mathscr{P}_i$ is nonprincipal so there is a $U_i \in \mathscr{P}_i$ with $x \notin U_i$ and $\{x\}$ and $\{i\} \cup U_i$ are the desired disjoint neighbourhoods.

The topology we obtain is totally disconnected. A connected subset $C$ of $X$ with 2 or more points cannot contain any points from the $\mathbb{N}_i$ since these points are open in $X$. Thus $C$ is a subspace of $Y := \mathbb{N} \cup \{p\}$. However, each point of $\mathbb{N}$ is open in $Y$, so none of these points can be in $C$ either and $C = \{p\}$ (contradiction).

Finally, the topology we obtain is not regular because we can't separate $p$ from the closed set $\mathbb{N}$.

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Dimitrije and Brian, I didn't see your answers until I was just about to post this, so I suspect this is now obsolete! – Mike F Nov 20 '11 at 7:32