Suppose $X$ is a totally disconnected compact Hausdorff space and $F_1,F_2,F_3$ three finite subsets of $X$.

I would like to know if $F_1 \cap F_2 \cap F_3 = \emptyset$ implies that there exists clopen sets $O_1, O_2, O_3$ such that $F_i \subseteq O_i$ and $O_1 \cap O_2 \cap O_3 = \emptyset$ ?

It seems false but I lack counterexample.

Edit : To find a counterexample, I think that a not discrete totally disconnected compact Hausdorff space which admits a finite clopen set should work (but I'm not still sure why and I'm not sure counterexamples exist).

  • $\begingroup$ You can separate all of the points in the finite sets and choose all of the separating open sets to be clopen and disjoint. What's the problem? $\endgroup$ – Robert Thingum Apr 3 at 18:19
  • $\begingroup$ I'm sorry, but could you give me an example for $F_1 = \lbrace x,y \rbrace$, $F_2 = \lbrace x,z \rbrace$ and $F_3 = \lbrace y,z \rbrace$ ? $\endgroup$ – L.DeR Apr 3 at 19:43

EDIT: My response does not answer the question in the OP. My response assumes that the finite sets are pairwise disjoint.

First you need to know the definition of the inductive dimension.

Definition Given a regular space $X$, the small inductive dimension $ind(X)$ is defined in the following way:

(a) $ind(X)=-1$ if and only if $X=\emptyset$

(b) For $n\geq 0$, $ind(X)\leq n$ if and only if for every $x\in X$ and every open neighbourhood $U\subseteq X$ of $x$ there is another open neighbourhood $V\subseteq X$ such that $x\in V\subseteq U$ and $ind(\partial V)\leq n-1$.

(c) $ind(X)=n$ if $ind(X)\leq n$ and $ind(X)\leq n-1$ does not hold.

(d) $ind(X)=\infty$ if $ind(X)\leq n$ does not hold for any $n$.

What is relevant for you is that $ind(X)=0$ if and only if every point has a neighbourhood basis of sets with empty boundaries, that is clopen sets.

The important fact you need is this:

Proposition: If a compact Hausdorff space $X$ is totally disconnected, then $ind(X)=0$.

This proposition appears in Nagami's "Dimension Theory".

Now let $X$ be a compact Hausdorff space that is totally disconnected. Let




These are closed compact sets in a compact Hausdorff space, so there are disjoint open sets $U_{1},U_{2}$ and $U_{3}$ such that $F_{i}\subseteq U_{i}$. Now, within a particular $U_{i}$, say $U_{1}$, you can find disjoint open sets $W_{1},\ldots, W_{n}$ such that $x_{i}\in W_{i}$. Because $ind(X)=0$ you can then pass to clopen sets $V_{1},\ldots,V_{n}$ such that $x_{i}\in V_{i}\subseteq W_{i}$. Then define $V=\bigcup_{i=1}^{n}V_{i}$. This is an open set containing $F_{1}$ and because the closures of the $V_{i}$ are all disjoint we have:

$$\partial V=\bigcup_{i=1}^{n}\partial V_{i}=\bigcup_{i=1}^{n}\emptyset=\emptyset$$

Thus $V$ is clopen and disjoint from $U_{2}$ and $U_{3}$. Repeat this process for $U_{2}$ and $U_{3}$ to get two more clopen sets.

  • $\begingroup$ I understand that if $F_1 \cap F_2 = F_1 \cap F_3 = F_2 = F_3 = \emptyset$, there exist disjoints clopen sets which answer my question. But what happen if $F_1 \cap F_2 \neq \emptyset$, $F_1 \cap F_3 \neq \emptyset$, $F_2 \cap F_3 \neq \emptyset$ but $F_1 \cap F_2 \cap F_3 = \emptyset$ ? How do you find the disjoint open sets $U_1, U_2$ and $U_3$ ? (and subsenquently, the disjoint clopen sets). $\endgroup$ – L.DeR Apr 3 at 20:34
  • $\begingroup$ Ah, I misunderstood your question, my apologies. In the case your speaking of I'm not sure. $\endgroup$ – Robert Thingum Apr 3 at 20:37
  • $\begingroup$ Not a problem, in the beginning I misunderstood the problem in the same way. $\endgroup$ – L.DeR Apr 3 at 20:54

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