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I am trying to expand my intuition of Boolean algebras and Stone's representation theorem. In particular I am trying to understand why the associated (Stone) space of a Boolean algebra is always Compact (It is also Hausdorff and Totally Disconnected). For clarity, I want to investigate the special case of a simple infinite Boolean Algebra and identify the apparent flaw in my line of reasoning. (Further generalizations and comments are welcome)

Let $B = \mathcal{P}(\omega)$ be the boolean algebra ordered by inclusion.

Let also $S(B)$ be the set of all utrafilters on $B$.

My understanding is that in our case this would mean that: $S(B) = \lbrace F_i \subseteq \mathcal{P}(\omega) \vert i \in \omega\rbrace$ where each $F_i$ is the principal ultrafilter on $B$ with $\lbrace i \rbrace$ being its minimal element.

We can induce a topology on $S(B)$ generated by the (basic open) sets $U_b=\lbrace F \in S(B) \vert b \in F \rbrace$

It seems to then follow that $U_{\lbrace i \rbrace} = F_i$ (for all $i \in \omega$) which would mean that all singletons of $S(B)$ are open.

But if that were the case, we simply have that $S(B)$ admits the dicrete topology on an infinite set, and thus it could not be Compact. This would seem to violate the fact that $S(B)$ should always be a Compact, Totally Disconnected, Hausdorff (Stone) space.

What am I missing?

UPDATE: It seems that I was missing quite a few ultrafilters in my calculation of $S(B)$. It was not particularly obvious to me as to why there should be any more than just the "obvious" principal ultrafilters. A step in the right direction (at least for me) was to realize that, informally speaking, every ultafilter contains "half" of the elements of the boolean algebra $B$. More formally, for every $b \in B$ and $F \in S(B)$ we have that either $b \in F$ or $\omega \setminus b \in F$ (which immediately means that the basic sets $U_b$ are also closed). The next - far from trivial - step is to observe that one can find several such "dichotomies" which form filters. As a matter of fact, the number of "filter-dichotomies" equals the number of the dichotomies themselves which is rather interesting.

Pellenthor
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$S(B)$ contains all countably many fixed ultrafilters $\mathcal{U}_n=\{A \subseteq \omega: n \in A\}$ (for any fixed $n \in \omega$)(also called "principal ultrafilters") but many many more ($2^{2^{\aleph_0}}$ many!) non-fixed, free ultrafilters, that have empty intersection, and no common point. These are hard to imagine (you cannot give a constructive example of one) but Zorn's lemma implies that they exist. The fixed ultrafilters as a subspace of $S(B)$ form a countable discrete set. The Stone space is homeomorphic to what topologists call the Cech-Stone (or Stone-Cech) compactification (see Wikipedia e.g.) of the integers and the fixed ultrafilters "are" in that compactfication the original integers. The other ultrafilters are the compactifying points we add.

So your picture is incomplete. K.P. has a nice intro to Stone duality here, as an extra resource.

Henno Brandsma
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  • What is a "fixed" ultrafilter? – Noah Schweber Jan 13 '20 at 22:55
  • @NoahSchweber one with non-empty intersection. As opposed to a free ultrafilter. Standard terms, so I thought. – Henno Brandsma Jan 13 '20 at 22:58
  • So "non-fixed" and "free" are synonyms? (Also, I've not seen that term before - do you know a source for it?) – Noah Schweber Jan 13 '20 at 22:59
  • @NoahSchweber for me, they are. – Henno Brandsma Jan 13 '20 at 23:00
  • @NoahSchweber "Discovering Modern Set Theory I", Just and Weese (p. 135) is the first I could find. – Henno Brandsma Jan 13 '20 at 23:05
  • @HennoBrandsma This is interesting thanks. I was worried I may be missing hidden ultrafilters, which would make sense, especially if the $AC$ is involved. The proof of the compactness of $S(B)$, is it non-trivial also? In general, what is the best intuitive way (if any) to look at $S(B)$ in order to extract all of its topological properties? – Pellenthor Jan 13 '20 at 23:13
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    @Pellenthor The compactness isn't that hard, see KP's notes I linked to. You only need to consider covers of $S(B)$ by basic elements, so that brings it back nicely to $B$ (as basic elements are determined by elements of $B$). It's a nice theory, Stone-duality. – Henno Brandsma Jan 13 '20 at 23:18
  • @HennoBrandsma Speaking very informally, and upward closure aside, does the $2^{2^{\aleph_0}}$ number of ultrafilters comes from the fact that for "almost all" elements $b \in \mathcal{P}(\omega)$ we have to make a (binary) choice of whether to include $b$ or $b'$ (its complement) in the ultrafilter? Is this why these extra ultrafilters have an empty intersection and are so hard to imagine? – Pellenthor Jan 14 '20 at 03:56
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    @Pellenthor yes, it’s a bit like that, yes. But there is a formal proof of the number on this site too, using independent families. Nice infinite combinatorics. – Henno Brandsma Jan 14 '20 at 05:05
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    @Pellenthor This nice answer by Brian W Scott gives an argument as to why there are so many free ultrafilters on an infinite set (works on any size infinite set, not just $\omega$). If you're curious.. – Henno Brandsma Jan 14 '20 at 07:02