Timeline for Are all free ultrafilters 'the same' in some sense?
Current License: CC BY-SA 4.0
16 events
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| Nov 24, 2023 at 13:31 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
deleted 3 characters in body
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| Feb 22, 2022 at 19:01 | answer | added | Taras Banakh | timeline score: 8 | |
| Dec 12, 2021 at 18:14 | comment | added | Kevin Buzzard | There is a sense in which most real numbers are the same -- noncomputable and "random". However the reals have trivial automorphism group. Perhaps the same is going on here. | |
| Dec 11, 2021 at 20:46 | vote | accept | Squala | ||
| Dec 11, 2021 at 2:44 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title, and other minor proofreading
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| Dec 11, 2021 at 2:37 | answer | added | markvs | timeline score: 10 | |
| Dec 10, 2021 at 20:59 | history | became hot network question | |||
| Dec 10, 2021 at 17:36 | comment | added | YCor | @NoahSchweber well, it's misleading if you ignore the "point" aspect. I believe one should have both aspects in mind. I still view, in this context, points to lie one level "deeper" than clopen subsets (= points in the corresponding BA). | |
| Dec 10, 2021 at 16:27 | comment | added | Noah Schweber | @YCor While technically true, I think that's highly misleading in this context since (even ignoring "pure" questions) there are important applications of "special" ultrafilters. | |
| Dec 10, 2021 at 15:07 | comment | added | YCor | I'd say yes in the following sense: for many "usual" proofs on ultrafilters, we often say "there is one nonprincipal ultrafilter such that", but one could say: "there is a nonempty clopen subset of the set of nonprincipal ultrafilters such that". And indeed, the action of the symmetric group on the set of nonempty proper clopen subsets of the space of nonprincipal ultrafilters is transitive. | |
| Dec 10, 2021 at 13:51 | answer | added | Will Brian | timeline score: 39 | |
| Dec 10, 2021 at 13:38 | comment | added | Asaf Karagila♦ | In some sense? Sure. All free ultrafilters on $\Bbb N$ have the same cardinality. | |
| Dec 10, 2021 at 13:38 | comment | added | Wojowu | The action of homeomorphisms on $\beta\mathbb N$ is far from transitive. Indeed, the isolated points of this space are precisely the principal ultrafilters, and they form a dense subspace, so the homeomorphism group is in bijection with permutations of $\mathbb N$ - meaning two ultrafilters are in the same orbit iff they are isomorphic (per the definition in your first paragraph). | |
| Dec 10, 2021 at 13:04 | comment | added | Benjamin Steinberg | $\beta \mathbb N$ has the structure of as compact right semitopological semigroup. The property of an ultrafilter being idempotent or a minimal idempotent is important in applications to Ramsey theory. | |
| S Dec 10, 2021 at 12:58 | review | First questions | |||
| Dec 10, 2021 at 13:18 | |||||
| S Dec 10, 2021 at 12:58 | history | asked | Squala | CC BY-SA 4.0 |