Are all minimal totally separated spaces compact? Let us call a space $(X,\tau)$ totally separated 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 totally separated.
Clearly compact totally separated spaces are minimal totally separated because they are minimal Hausdorff, as a basic theorem of general topology asserts. Conversely, is every minimal totally separated space compact?
 A: 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 $A\to B$. 
