Timeline for Why are extremally disconnected spaces so hard to give examples of?
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17 events
| when toggle format | what | by | license | comment | |
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| Sep 19 at 16:48 | answer | added | Ramiro de la Vega | timeline score: 3 | |
| Aug 16 at 14:26 | comment | added | Will Sawin | @Gro-Tsen Actually, if we have an explicit non-principal ultrafilter on $S$, we can define a topology on $S \cup \{\infty\}$ where a set is open unless it contains $\infty$ but its restriction to $S$ does not lie in the ultrafilter. So this really gives as an example that is as explicit as possible under AD. | |
| Aug 16 at 10:21 | comment | added | Gro-Tsen | @WillSawin I think I can simplify your example a bit: by a result of Solovay (that is related to the one you mention: see here), AD implies that the club filter on $ω_1$ (i.e., the filter generated by all closed unbounded subsets of $ω_1$) is an ultrafilter. So this provides an “explicit” point of the Stone-Čech compactification of the discrete topology on $ω_1$ (NB: not to be confused with the S-Č compactification of the usual topology on $ω_1$, which is just $ω_1 + 1$, but the latter is not extremally disconnected). | |
| Aug 16 at 0:06 | comment | added | Will Sawin | Here is a strange example. We can consider the Stone space of the Boolean algebra of subsets of the sets of Turing degrees (inspired by Joseph Van Name's discussion of the relation to non-principal ultrafiliters) Under the axiom of determinacy, the Martin measure en.wikipedia.org/wiki/Martin_measure gives an explicit non-principal ultrafilter and thus an explicit point of this space. So this gives an example that is in one sense explicit - one interesting point is explicit, even if it's not clear whether other points are explicit - under an assumption very far from the axiom of choice. | |
| Aug 15 at 20:51 | comment | added | Gro-Tsen | @godelian Thanks: indeed, I can read the deleted answer, and it's still instructive. | |
| Aug 15 at 20:28 | comment | added | godelian | @Gro-Tsen I wrote the example but as pointed out by Will Sawin in a comment it is only Hausdorff when the space is discrete, thus uninteresting. The deleted answer is still visible to you if you want to check it out. | |
| Aug 15 at 19:56 | comment | added | Joseph Van Name | If $\kappa$ is a measurable cardinal and $\mathcal{U}$ is a normal ultrafilter on $\kappa$, then the collection $[\kappa]^{<\omega}$ can be given a non-discrete extremally disconnected topology where if $U\subseteq [\kappa]^{<\omega}$ is open if and only if whenever $A\in U$, there is some $R\in\mathcal{U}$ with $\max(A)<\min(R)$ and where if $A\subseteq B\subseteq A\cup R$, $|B|<\omega$, then $B\in U$ as well. The points of this space are explicit, and the normal ultrafilter on $\kappa$ is unique in some models of set theory. | |
| Aug 15 at 18:07 | comment | added | Gro-Tsen | @godelian This seems promising, and your example interests me. I can (sort of) form a mental picture of an $η_1$ totally ordered set, like the set of functions $ω_1 \to \{-1,0,+1\}$ that are eventually 0, ordered lexicographically, and I think $η_1$ means the same as $\aleph_1$-saturated for the order structure. Can you describe what the topology you are referring to is, more concretely, in this example (e.g., how it relates to the order topology), which would make it extremally disconnected? (Please don't hesitate to make an answer out of this.) | |
| Aug 15 at 17:12 | comment | added | Gerald Edgar | Is the existence of non-discrete examples independent of ZF? "Requires axiom of choice" is one version of "not explicit". | |
| Aug 15 at 16:47 | comment | added | mathworker21 | @DenisT What does "ind-finite" mean? | |
| Aug 15 at 16:34 | history | became hot network question | |||
| Aug 15 at 15:00 | answer | added | Joseph Van Name | timeline score: 20 | |
| Aug 15 at 8:10 | comment | added | Alessandro Codenotti | Every extremally disconnected Hausdorff space $X$ is a subspace (retract if $X$ is compact) of $\beta Y$ for some discrete $Y$, so they all seem to be ultrafilter-ish, even though the above doesn't exclude the possibility of an extremally disconnected space defined with no explicit mentions of ultrafilters | |
| Aug 15 at 1:10 | comment | added | Denis T | One very reasonable argument for them being 'hard to visualise' is that those spaces cannot have countable base of topology; and everything 'visualise-able' ought to be ind-finite over a countable diagram (as far as my knowledge of 'visualising' goes, according to me and a few people who I talked to about this). | |
| Aug 14 at 23:46 | comment | added | godelian | If you take the underlying set of a $\kappa^+$-saturated model and consider the topology whose opens are the subsets definable by $\kappa$-ary positive existential formulas then the resulting space is extremally disconnected. Is that the kind of concrete example you were looking for? | |
| Aug 14 at 23:23 | comment | added | Ali Taghavi | With a Non commutative point of view whose world consits of point less space they are not rare: every commutative von neumann algebra is an example of an extremely disconnected space: Consider its spectrum | |
| Aug 14 at 23:14 | history | asked | Gro-Tsen | CC BY-SA 4.0 |