Intereting Posts

Prove that $\sum_{r=1}^n \frac 1{r}\binom{n}{r} = \sum_{r=1}^n \frac 1{r}(2^r – 1)$
Is there a “natural” topology on powersets?
How to compute coefficients in Trinomial triangle at specific position?
List of interesting integrals for early calculus students
Proof: Show there is set of $n+1$ points in $\mathbb{R}^n$ such that distance between any two distinct points is $1$?
What is the number of bijections between two multisets?
Maximal ideals in polynomial rings
Differentiation with respect to a matrix (residual sum of squares)?
Distance between a point and closed set in finite dimensional space
Meaning of variables and applications in lambda calculus
Showing a ring where $ax = b$ has a solution for all non-zero $a, b$ is a division ring
Open set as a countable union of open bounded intervals
Proof that two spaces that are homotopic have the same de Rham cohomology
Why is $L_A$ not $\mathbb K$ linear (I can prove that it is)
Self-teaching myself math from pre-calc and beyond.

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.

- Must a weakly or weak-* convergent net be eventually bounded?
- Is the outer boundary of a connected compact subset of $\mathbb{R}^2$ an image of $S^{1}$?
- Connected, locally connected, path-connected but not locally path-connected subspace of the plane
- Connectedness problem: sequences of points with distances at most $\varepsilon$
- Discontinuous functions with closed graphs
- Intersection of two open dense sets is dense

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:

- $U \cap \mathbb{Q}$ is open in the standard topology on $\mathbb{Q}$.
- If $p_0 \in U$, then $U$ contains all but finitely many of $I_0,I_2,I_4,\ldots$
- 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.

- Are these two definitions equivalent?
- Intuition of the meaning of homology groups
- Proof for “Given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size.”
- The general argument to prove a set is closed/open
- separation properties in Hausdorff, compact spaces
- every topological space can be realized as the quotient of some Hausdorff space.
- Lebesgue measure paradox
- Intersection and union of simply connected domain
- Prove that the unit open ball in $\mathbb{R}^2$ cannot be expressed as a countable disjoint union of open rectangles.
- Why are the fundamental groups $\pi_{1}(S^3)$ and $\pi_{1}(S^2)$ trivial?

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.

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.

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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