2
$\begingroup$

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?

$\endgroup$
10
  • $\begingroup$ Dominic, in your two recent posts (Questions, including this one) you are using your own definition of the 0-dimensional space which drastically contradicts the well established classical topological terminology. You should add a triple X rating so that mathematicians under 18 will not be exposed to it. Other people could go on, they worked on your questions--I couldn't. Regards. $\endgroup$ Commented Apr 18, 2015 at 15:45
  • 2
    $\begingroup$ Wikipedia suggests this property (the one you call "zero-dimensional") already has a name: it's called "totally separated". en.wikipedia.org/wiki/Connected_space#Disconnected_spaces $\endgroup$ Commented Apr 18, 2015 at 17:15
  • $\begingroup$ OK thanks for correcting my terminology, I promise to adhere to it from now on. Will change my posts on Monday, when I'm online again. $\endgroup$ Commented Apr 18, 2015 at 20:02
  • 2
    $\begingroup$ In the context of this question, it seems to me that "totally separated" and "zero-dimensional" are not that different. Namely, a minimal totally separated topology is zero-dimensional (it's generated by its clopen sets), so "minimal totally separated" is the same as "minimal zero-dimensional Hausdorff". Right? $\endgroup$
    – bof
    Commented Apr 19, 2015 at 0:50
  • 2
    $\begingroup$ It looks like this question has been answered by Eric Wofsey here mathoverflow.net/a/203252/22277 in a similar question posted. $\endgroup$ Commented Apr 19, 2015 at 20:24

1 Answer 1

1
$\begingroup$

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 $A\to B$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .