Timeline for What is your favorite proof of Tychonoff's Theorem?
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2021 at 12:14 | comment | added | Pierre-Yves Gaillard | @ACL - I see that I hadn't answered your comment. Sorry! I don't remember the details, but I was referring to a paragraph on p. 38 of Weil's book on topological groups. I added a screenshot of the paragraph to the question. I'm afraid I overstated Weil's comment. | |
| Nov 19, 2012 at 15:12 | comment | added | ACL | @Pierre-Yves: Do you know what Weil could mean with a proof of the existence of a Haar measure where filters are not hidden? Thank you in advance. | |
| Oct 13, 2012 at 4:04 | vote | accept | Pierre-Yves Gaillard | ||
| May 7, 2017 at 10:41 | |||||
| Jan 28, 2011 at 7:53 | history | edited | Pietro Majer | CC BY-SA 2.5 |
deleted 10 characters in body
|
| Jun 1, 2010 at 3:39 | comment | added | Pierre-Yves Gaillard | Loomis's proof could be dubbed "Tychonoff's Theorem for dummies". That's why I like it! | |
| May 31, 2010 at 18:48 | comment | added | Pietro Majer | Dear Pierre-Yves, actually, I think I would profit of Loomis' presentation (thank you btw) in the case of a short course on a specialized topic, with few time for detours, and where Tychonoff theorem would have an important role (assuming it were for some reason unknown to the audience). I would also give references for other proofs. | |
| May 30, 2010 at 10:50 | comment | added | Pierre-Yves Gaillard | Dear Pietro, Pete, Spencer: In "L'intégration dans les groupes topologiques" Weil explains that his proof of the existence of a Haar measure on locally compact groups was obtained by hiding filters. I think the situations are very similar. (And I find Weil's proof incredibly beautiful - for the ideas and the style.) | |
| May 30, 2010 at 9:58 | comment | added | Spencer | Indeed Michael. I am not one to worry about the axiom of choice usually (i.e. it's use wasn't my main point) and probably as a result of this I do remember a struggle with some friends to find where we had used full choice in said proof, since it was just one word: blah blah blah ....then F converges to x, say...... | |
| May 30, 2010 at 9:14 | comment | added | Pierre-Yves Gaillard | (Cont.) I know that Hecke was a much greater mathematician than I, but if had to teach this topic, I would still give an elementary proof. | |
| May 30, 2010 at 9:14 | comment | added | Pierre-Yves Gaillard | Dear Spencer and Pietro Majer: I have no argument against your point, but I still prefer Loomis's phrasing. (I'd lie if I said the contrary.) I often find myself in this kind of situation. The (unrelated) example that strikes me most is Hecke's "Lectures on the Theory of Algebraic Numbers", where he says: There are elementary proofs of the quadratic reciprocity, but "they possess rather the character of supplementary verification". So "we will dispense entirely with a presentation of an elementary proof". (Cont.) | |
| May 30, 2010 at 9:11 | comment | added | Pete L. Clark | I also like Cartan's ultrafilter proof best. As people have said, all the cleverness is hidden in the definition of ultrafilters. With this in hand, Tychonoff's theorem becomes a consequence of a few very straightforward facts about filters on products. As to whether it's better to hide the ultrafilters or not, I think the test should be whether ultrafilters are of any use outside of this specific context, the answer to which is of course YES! Thus I would recommend this proof even to someone who doesn't already know about ultrafilters -- learning about them is good in and of itself. | |
| May 30, 2010 at 8:55 | comment | added | Michael Greinecker | Actually, the axiom of choice is used twice in the proof. First, you have to use it to characterize compactness by ultrafilters. For this, you do not need the full axiom of choice, the boolean prime ideal theorem suffices. The second application is in picking for each coordinate a point the projection converges to. Here you need the full axiom of choice. For Hausdorff spaces, you don't need the second part though, because than a filter cannot converge to different points. | |
| May 30, 2010 at 8:38 | comment | added | Spencer | [My previous is in response to Pierre's] | |
| May 30, 2010 at 8:37 | comment | added | Spencer | I think perhaps it's more accurate to say that the ultrafilters hide the details of the Loomis proof. The fiddling around with FIP and Zorn's and what not goes into setting up ultrafilters and characterizing compactness, after which the proof is two-three lines long. | |
| May 30, 2010 at 8:28 | comment | added | Pietro Majer | After a closer look at Loomis' one, I agree: in the substance it's the same as H.Cartan (it seems to me your're right with the attribution to Cartan, but I'm not certain either). But, I think here's an important issue: to give or not to give a name to the objects one uses? Sometimes there is no need at all (you know those papers with definitions of weird objects that only enter once in a proof). In this case, I think the proof greatly gains in semplicity introducing the notion of filter. I would even say, the notion of filter is the most important byproduct of the compactness theorem. | |
| May 30, 2010 at 7:58 | comment | added | Pierre-Yves Gaillard | Dear Pietro Majer: I think Loomis's proof (the one I gave) is the same as the one you mention (due to H. Cartan if I'm not mistaken). Loomis hides the (ultra)filters. But I think he hides them very nicely. (Loomis credit the proof he gives to Bourbaki.) | |
| May 30, 2010 at 7:40 | comment | added | Spencer | Yes. I can't say I know lots of proofs of Tychonoff's, but I think that the ultrafilter proof is very nice indeed. Property 2. you mention seems to show very clearly how they are a useful generalization of sequences. | |
| May 30, 2010 at 7:37 | history | edited | Pietro Majer | CC BY-SA 2.5 |
added 102 characters in body
|
| May 30, 2010 at 7:25 | history | answered | Pietro Majer | CC BY-SA 2.5 |