Timeline for What is your favorite proof of Tychonoff's Theorem?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 22, 2022 at 20:07 | comment | added | Todd Trimble | @StefanWitzel One makes a choice at each successor stage, but you're right that AC as such is invoked at limit stages by assembling those infinite many prior choices. I don't think you're missing anything, and thanks for promoting clarity. | |
| May 20, 2022 at 11:57 | comment | added | Stefan Witzel | @ToddTrimble: You say there is choice at successor stages, but that's just picking an element, right? It seems to me the actual choice (apart from the well-ordering) is in the limit step when you go from "there exists a bunch of $x_\beta$" to "there is a tuple $(x_\beta)_{\beta < k}$". Or am I missing something? | |
| Nov 27, 2021 at 7:10 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https
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| May 7, 2017 at 12:40 | comment | added | Todd Trimble | No problem, Pierre-Yves! I agree that it's artificial to have to choose (I'm old-fashioned and really like the ultrafilter-type proofs for their overall conceptual simplicity). "L'embarras du choix" :-) | |
| May 7, 2017 at 12:31 | comment | added | Pierre-Yves Gaillard | @ToddTrimble - Thanks Todd for this nice answer. I unaccepted it because I accepted it by mistake. I just wanted to upvote it on my iPad and didn't put my finger at right spot. Sorry! Again, I like your answer very much, but there are so many great answers that it would seem somewhat artificial to me say that one of them is better than the others. | |
| May 7, 2017 at 12:29 | comment | added | Todd Trimble | There is choice in choosing a lift at successor stages. I think the only applications of AC are that and the well-ordering of the indexing. The characterization of compactness in terms of closed projections doesn't require AC and is something you can enact in bounded Zermelo set theory. For example, you don't need to invoke nonprincipal ultrafilters there. | |
| May 7, 2017 at 11:24 | comment | added | David Roberts♦ | And this proof has the nice feature, it seems, of proving the theorem for well-ordered families in situations without Choice. Or are there other appeals to some choice principle in the lemma or the characterisation of compactness? Clearly getting the full theorem for arbitrary families from the above requires the well-ordering principle, as one would expect... | |
| May 7, 2017 at 11:13 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added some words for clarity
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| May 7, 2017 at 10:41 | vote | accept | Pierre-Yves Gaillard | ||
| May 7, 2017 at 12:21 | |||||
| May 7, 2017 at 0:43 | history | edited | Todd Trimble | CC BY-SA 3.0 |
cardinal --> ordinal
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| S May 7, 2017 at 0:21 | history | answered | Todd Trimble | CC BY-SA 3.0 | |
| S May 7, 2017 at 0:21 | history | made wiki | Post Made Community Wiki by Todd Trimble |