Intereting Posts

Properties of special rectangle (measure)
Covariant derivative in $\mathbb{R}^n$
Sum of two open sets is open?
Is this proof correct? proving a sum of a convergent and divergent sequences is a divergent sequence
integrate $\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$
Conic by three points and two tangent lines
Is an irreducible element still irreducible under localization?
L-functions identically zero
About the computation of the limit $ \lim_{x \to \infty} \frac{1^{99} + 2^{99} + \cdots + x^{99}}{x^{100}} $
Area of a cyclic polygon maximum when it is a regular polygon
Determinant of a non-square matrix
Using the definition of limit prove the following
Is there an infinite connected topological space such that every space obtained by removing one point from it is totally disconnected?
On the commutative property of multiplication (domain of integers, possibly reals)
Any two norms equivalent on a finite dimensional norm linear space.

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.

- Tietze extension theorem for complex valued functions
- A homomorphism induces a continuous map from ${\rm Spec}(A') \to {\rm Spec}(A)$.
- Difference between two definitions of Manifold
- every non-principal ultrafilter contains a cofinite filter.
- Prove: $f: \mathbb{R} \rightarrow \mathbb{R}$ st for every $x \in \mathbb{R}$ there exists $n$ st $f^{(n)}(x) = 0$, f is a polynomial.
- Are Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?

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.

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- Intersection of a closed set and compact set is compact
- About the equivalence of definitions of a Baire Space
- book with lot of examples on abstract algebra and topology
- Understanding the definition of continuity between metric spaces.
- Can continuity be proven in terms of closed sets?

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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