# Minimal totally separated spaces

Let us call a space $(X,\tau)$ totally separated (ts) if for every two distinct points there is a clopen set containing one, but not the other. If for every topology $\sigma\subseteq\tau$ with $\sigma\neq \tau$ the space $(X,\sigma)$ no longer has this property we call $(X,\tau)$ minimal ts.

If $(X,\tau)$ is ts, does $\tau$ contain a topology $\sigma$ such that $(X,\sigma)$ is minimal ts?

• Your definition of "zero-dimensional" is actually known as totally disconnected. Zero-dimensionality is usually defined as "has a basis consisting of clopen sets". – Andrej Bauer Apr 18 '15 at 13:32
• For small inductive dimension, zero-dimensionality is equivalent to "clopens form a basis" (see Wikipedia if you trust it, en.wikipedia.org/wiki/Zero-dimensional_space). A space is totally disconnected iff its components are singletons, see en.wikipedia.org/wiki/Totally_disconnected_space. This is equivalent to saying that every two points can be separated by a clopen, which is what you wrote. – Andrej Bauer Apr 18 '15 at 14:07
• @AndrejBauer The wiki article you quote about totally disconnected spaces implies these spaces are $T_1$ but not necessarily Hausdorff. My definition in the original post implies Hausdorffness. – Dominic van der Zypen Apr 18 '15 at 14:12
• It does not matter. There are Hausdorff spaces which match your definition ("clopens separate"), but they are not $0$-dimensional in the standard sense (i.e., small inductive dimension). Is there a standard notion of dimension for which your definitions means "$0$-dimensional"? – Andrej Bauer Apr 18 '15 at 14:15
• @DominicvanderZypen: An easy way to get a non-$T_2$ totally disconnected space is to take a totally disconnected space and double a non-isolated point. – Eric Wofsey Apr 18 '15 at 14:27

First, note that a minimal totally separated space is the same thing as a Stone space. Clearly Stone spaces are minimal totally separated (any coarser topology cannot even be Hausdorff); conversely suppose $X$ is totally separated and not Stone. We may assume the topology on $X$ is generated by its clopen sets (otherwise they generate a coarser totally separated topology). Then $X$ is canonically a dense subspace of the Stone space $S(B)$ of its clopen algebra $B$. If $X$ is not all of $S(B)$, let $u\in S(B)\setminus X$ and $x\in X$. Let $T$ be the quotient of $S(B)$ obtained by identifying $x$ and $u$; the composition $X\to S(B)\to T$ is then injective and induces another totally separated topology on $X$. This new topology is strictly coarser than the original topology: there is some net $(x_i)$ in $X$ that converges to $u$ in $S(B)$, and this net (which had no limit in $X$ in the old topology) converges to $x$ in the new topology.

Thus a minimal totally separating topology contained in a given topology on $X$ is equivalent to a continuous bijection $X\to S$ from $X$ to a Stone space $S$. If $A$ is the clopen algebra of $S$, then $A$ is naturally a subalgebra of the clopen algebra $B$ of $X$, and the map $X\to S$ is determined by the inclusion $A\to B$. Thus the question can be recast as follows: given a totally separated space $X$ with clopen algebra $B$, is there a subalgebra $A\subseteq B$ such that the canonical map $X\to S(A)$ to the Stone space of $A$ is a bijection?

Let $D$ be any infinite discrete space; then I claim we can find a counterexample $X$ which is a dense subspace of the Stone-Cech compactification $\beta D$. To find such an $X$, note that $|\beta D|=2^{2^{|D|}}$, which is the same as the number of subalgebras of the power set algebra $\mathcal{P}(D)$. We can thus by transfinite induction build a subset $X\subset \beta D$ that contains $D$ and avoids bijecting onto $S(A)$ for each subalgebra $A\subseteq \mathcal{P}(D)$, identifying $\mathcal{P}(D)$ with the clopen algebra of $\beta D$. Since $X$ contains $D$ as a dense subset, a clopen subset of $X$ is determined by its intersection with $D$, and so $\mathcal{P}(D)$ is also the clopen algebra of $X$. By construction, then, the map $X\to S(A)$ is not a bijection for any subalgebra $A$ of the clopen algebra of $X$.

This is, of course, horribly nonconstructive. It would be interesting to see an explicit counterexample.

• Why are you allowed to make the assumption that the topology of $X$ is generated by the clopen sets? – Andrej Bauer Apr 18 '15 at 14:04
• Oh I see, you're saying you'll find an example which is not only totally disconnected (what OP called "0d") but is also zero-dimensional (clopens form a basis). – Andrej Bauer Apr 18 '15 at 14:09
• I just meant that for the purposes of the question, you might as well assume clopens form a basis, since if they don't then the topology they generate is a smaller 0d topology. – Eric Wofsey Apr 18 '15 at 14:11
• Ah, excellent. And so your last remark also shows that we do not even have to do anything with Stone spaces to find a counter-example: just take a totally disconnected space which is not $0$-dimensional. – Andrej Bauer Apr 18 '15 at 14:12
• @AndrejBauer: I don't understand your last comment. (Also, thanks for the edit :) ). – Eric Wofsey Apr 18 '15 at 14:22