Timeline for What is your favorite proof of Tychonoff's Theorem?
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11 events
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| May 7, 2017 at 17:52 | comment | added | Todd Trimble | I thought that Kelley's book (General Topology) had a correct proof of Tychonoff implies AC: let $\{X_i\}_{i \in I}$ be a family of nonempty sets. Put $Y_i = X_i \sqcup \{p\}$. Topologize $Y_i$ by taking the nontrivial open sets to be $X_i$ and $\{p\}$. Then $Y_i$ is compact; by Tychonoff, $Y = \prod_i Y_i$ is compact. For each $i$, put $K_i = \{y \in Y: y_i \in X_i\}$. Then $K_i$ is closed, and any finite intersection of the $K_i$ is nonempty (use $p$ in all but finitely many components). Hence $\prod_i X_i = \bigcap_\alpha K_i$ is nonempty as well, by compactness. Thus AC follows. | |
| May 30, 2010 at 13:59 | comment | added | Michael Greinecker | As Georges Elencwajg points out, there is a minor flaw in the original proof by Kelley that Tychonoff implies AC. Kelley uses the fact that the product of cofinite topologies is compact, a weaker statement equivalent to the theorem that every filter can be extended to an ultrafilter as shown here: math.vanderbilt.edu/~schectex/papers/kelley.pdf One can easily correct the original proof by Kelley by adding {Lambda} to the open sets, so the original proof is easily corrected. | |
| May 30, 2010 at 8:58 | history | edited | Dmitri Pavlov | CC BY-SA 2.5 |
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| May 30, 2010 at 8:48 | comment | added | Pierre-Yves Gaillard | Dear John: In Bourbaki's theory the "axiom of choice" is built-in. If you remove it, everything falls down. (You even loose the definition of the quantifiers.) The fact that a product of nonempty sets is nonempty is obvious (in this theory), and "Zorn's Lemma" is a theorem. (It would be interesting to know if people like Serre and Grothendieck have ever used the expression "axiom of choice" (in their writings).) | |
| May 30, 2010 at 8:24 | comment | added | Georges Elencwajg | Dear John, thanks for the link to Kelley's article. 1) It is interesting to note that Kelley endows the given sets with the Zariski topology: when was the article published? [Anyway I can imagine that people knew the Zariski topology under another name] 2) Kelley claims on page 76, line 11, of the linked article: "Surely $Z_a$ is closed...since $X_a$ is closed in $Y_a$." I think this is completely false if $X_a$ is infinite. Am I missing something? | |
| May 30, 2010 at 8:19 | comment | added | John Stillwell | @Pierre-Yves. I'm glad you like the Dirichlet translation; sorry about the axiom of choice (I don't know an equally nice proof that Tychonoff implies Zorn, but there may be one). | |
| May 30, 2010 at 7:53 | comment | added | Pierre-Yves Gaillard | Dear Amadeus: I had the impression that Munkres's proof was almost the same as the Loomis's (the one I gave), which was written long before. (Thank you for correcting me if I'm wrong.) | |
| May 30, 2010 at 7:42 | comment | added | Pierre-Yves Gaillard | Dear John Stillwell: As a bourbakist, the expression "axiom of choice" makes no sense to me. But I thank you very much for your comment. (Unrelated aside: Thank you very much also for your wonderful translation of Dirichlet! It changed my life!) Thank you to all contributors! | |
| May 30, 2010 at 6:35 | comment | added | Henno Brandsma | The original proof used the characterization (of compactness) that every infinite set has a point of complete accumulation, and involved a transfinite recursion, IIRC. It's probably in Fundamenta as well. | |
| May 30, 2010 at 6:31 | comment | added | John Stillwell | Every proof uses the axiom of choice, because Tychonoff is equivalent to AC. For the converse, that Tychonoff's theorem implies the axiom of choice, it's hard to beat the original proof by Kelley in Fundamenta Mathematicae 37 (1950) 75--76. It may be viewed here: matwbn.icm.edu.pl/ksiazki/fm/fm37/fm3716.pdf | |
| May 30, 2010 at 6:15 | history | answered | Amadeus | CC BY-SA 2.5 |