What is your favorite proof of Tychonoff's Theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:06:08Z http://mathoverflow.net/feeds/question/26416 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem What is your favorite proof of Tychonoff's Theorem? Pierre-Yves Gaillard 2010-05-30T04:44:19Z 2012-11-19T12:04:33Z <p>Here is mine. It's taken from page 11 of "An Introduction To Abstract Harmonic Analysis", 1953, by Loomis: </p> <p><a href="http://www.archive.org/details/introductiontoab031610mbp" rel="nofollow">http://www.archive.org/details/introductiontoab031610mbp</a> </p> <p><a href="http://ia331316.us.archive.org/3/items/introductiontoab031610mbp/introductiontoab031610mbp.pdf" rel="nofollow">http://ia331316.us.archive.org/3/items/introductiontoab031610mbp/introductiontoab031610mbp.pdf</a> </p> <p>(By the way, I don't know why this book is not more famous.) </p> <p>To prove that a product $K=\prod K_i$ of compact spaces $K_i$ is compact, let $\mathcal A$ be a set of closed subsets of $K$ having the finite intersection property (FIP) --- <em>viz.</em> the intersection of finitely many members of $\mathcal A$ is nonempty ---, and show $\bigcap\mathcal A\not=\varnothing$ as follows.</p> <p>By Zorn's Theorem, $\mathcal A$ is contained into some maximal set $\mathcal B$ of (not necessarily closed) subsets of $K$ having the FIP.</p> <p>The $\pi_i(B)$, $B\in\mathcal B$, having the FIP and $K_i$ being compact, there is, for each $i$, a point $b_i$ belonging to the closure of $\pi_i(B)$ for all $B$ in $\mathcal B$, where $\pi_i$ is the $i$-th canonical projection. It suffices to check that $\mathcal B$ contains the neighborhoods of $b:=(b_i)$. Indeed, this will imply that the neighborhoods of $b$ intersect all $B$ in $\mathcal B$, hence that $b$ is in the closure of $B$ for all $B$ in $\mathcal B$, and thus in $A$ for all $A$ in $\mathcal A$.</p> <p>For each $i$ pick a neighborhood $N_i$ of $b_i$ in such a way that $N_i=K_i$ for almost all $i$. In particular the product $N$ of the $N_i$ is a neighborhood of $b$, and it is enough to verify that $N$ is in $\mathcal B$. As $N$ is the intersection of finitely many $\pi_i^{-1}(N_i)$, it even suffices, by maximality of $\mathcal B$, to prove that $\pi_i^{-1}(N_i)$ is in $\mathcal B$.</p> <p>We have $N_i\cap\pi_i(B)\not=\varnothing$ for all $B$ in $\mathcal B$ (because $b_i$ is in the closure of $\pi_i(B)$), hence $\pi_i^{-1}(N_i)\cap B\not=\varnothing$ for all $B$ in $\mathcal B$, and thus $\pi_i^{-1}(N_i)\in\mathcal B$ (by maximality of $\mathcal B$). </p> <p>A pdf version is available at <a href="http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Tycho/" rel="nofollow">http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Tycho/</a> .</p> <hr> <p>Many people credit the general statement of Tychonoff's Theorem to Cech. But, as pointed out below by KP Hart, Tychonoff's Theorem seems to be entirely due to ... Tychonoff. This observation was already made on page 636 of</p> <p>Chandler, Richard E.; Faulkner, Gary D. Hausdorff compactifications: a retrospective. Handbook of the history of general topology, Vol. 2 (San Antonio, TX, 1993), 631--667, Hist. Topol., 2, Kluwer Acad. Publ., Dordrecht, 1998</p> <p><a href="http://books.google.com/books?id=O2Hwaj2SqigC&amp;lpg=PA636&amp;ots=xjvA9nwlO5&amp;dq=772%20tychonoff&amp;pg=PA636#v=onepage&amp;q&amp;f=false" rel="nofollow">http://books.google.com/books?id=O2Hwaj2SqigC&amp;lpg=PA636&amp;ots=xjvA9nwlO5&amp;dq=772%20tychonoff&amp;pg=PA636#v=onepage&amp;q&amp;f=false</a></p> <p>The statement is made by Tychonoff on p. 272 of "Ein Fixpunktsatz"</p> <p><a href="http://www.springerlink.com/content/n61706447r886l58/?p=328f0106a2634abfb53531fca0ca5a90&amp;pi=0" rel="nofollow">http://www.springerlink.com/content/n61706447r886l58/?p=328f0106a2634abfb53531fca0ca5a90&amp;pi=0</a></p> <p>where he says that the proof is the same as the one he gave for a product of intervals in "Über die topologische Erweiterung von Räumen"</p> <p><a href="http://www.springerlink.com/content/l656352441w67612/?p=328f0106a2634abfb53531fca0ca5a90&amp;pi=1" rel="nofollow">http://www.springerlink.com/content/l656352441w67612/?p=328f0106a2634abfb53531fca0ca5a90&amp;pi=1</a></p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26422#26422 Answer by Henno Brandsma for What is your favorite proof of Tychonoff's Theorem? Henno Brandsma 2010-05-30T05:26:43Z 2010-05-30T06:36:36Z <p>I like the proof from Alexander's subbase lemma. E.g. <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2005;task=show_msg;msg=1747" rel="nofollow">A proof here</a>. That lemma also gives the compactness criterion in ordered spaces (completeness implies compactness).</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26425#26425 Answer by Amadeus for What is your favorite proof of Tychonoff's Theorem? Amadeus 2010-05-30T06:15:20Z 2010-05-30T08:58:11Z <p>I first learnt from Munkres' Topology. He gave a different motivation to use the maximal principle (Hausdorff's to be precise, but Zorn's work too) instead of the historic motivation to characterize compact spaces with a generalized version of "sequence"; i.e. filters.</p> <p>What was Tychonoff's original proof? To me every proof seem to use some maximal principle; Alexander's subbase theorem also uses Zorn's lemma.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26426#26426 Answer by Andrew L for What is your favorite proof of Tychonoff's Theorem? Andrew L 2010-05-30T06:49:28Z 2010-05-30T06:49:28Z <p>My favorite is the proof via nets by Paul Chernoff. A VERY clever use of generalized convergence in point set topology! <a href="http://www.jstor.org/pss/2324485" rel="nofollow">http://www.jstor.org/pss/2324485</a></p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26427#26427 Answer by Andrej Bauer for What is your favorite proof of Tychonoff's Theorem? Andrej Bauer 2010-05-30T06:55:12Z 2010-05-30T07:02:38Z <p>My favorite proof is the one from Johnstone's Stone spaces for locales because it works without the axiom of choice.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26430#26430 Answer by Pietro Majer for What is your favorite proof of Tychonoff's Theorem? Pietro Majer 2010-05-30T07:25:28Z 2011-01-28T07:53:29Z <p>Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC) </p> <blockquote> <p>X is compact if and only if every ultrafilter is convergent.</p> </blockquote> <p>Then one observes that </p> <ol> <li><p>any image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)</p></li> <li><p>any filter in the product space converges if and only if all its projections converge .</p></li> </ol> <p>You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car? </p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/26454#26454 Answer by KP Hart for What is your favorite proof of Tychonoff's Theorem? KP Hart 2010-05-30T14:41:20Z 2010-06-01T07:58:39Z <p><a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002273799" rel="nofollow">Here</a> is Tychonoff's original proof, for powers of the unit interval. He builds a complete accumulation point of a given infinite set by transfinite recursion along the index set. On page 772 of <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002278103" rel="nofollow">this paper</a> one finds the formulation of the general theorem (in my translation): "The product of compact spaces is again compact. One proves this theorem word for word as in he case of the compactness of the product of intervals". Some authors (Folland, see comment below and Walter Rudin in his `Functional Analysis') credit &#268;ech with proving the general result but &#268;ech's proof is the same as Tychonoff's and, based on a reading of his papers, I think Tychonoff deserves full credit for the theorem and its proof. </p> <p>@Henno: not Fundamenta but Mathematische Annalen.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/53584#53584 Answer by Michael Blackmon for What is your favorite proof of Tychonoff's Theorem? Michael Blackmon 2011-01-28T09:09:42Z 2011-01-28T09:09:42Z <p>Personally, I've always enjoyed the proof given in Topology, by Hocking and Young. It's essentially the basic ultrafilter proof, but its got a nice feel to it. I guess I'm biased because this was the first real Topology book I was ever able to get my hands on.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/53586#53586 Answer by J.J. Green for What is your favorite proof of Tychonoff's Theorem? J.J. Green 2011-01-28T09:28:48Z 2011-01-28T09:28:48Z <p>A very short proof using nonstandard analysis in M. Machover, J.L. Bell, A Course in Mathematical Logic (1977), </p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/65510#65510 Answer by jasomill for What is your favorite proof of Tychonoff's Theorem? jasomill 2011-05-20T08:39:03Z 2011-05-20T08:39:03Z <p>The non-standard analysis proof is an interesting "application" of the ultrafilter proof: a topological space $A$ is compact if and only if every point in the associated "non-standard topological space" ${}^*A$ is <em>near-standard</em>, that is to say, if and only if each $x \in {}^*A$ is contained in every open neighborhood of some standard point $y \in A$ ($i.e.,$ for all $U \subset A$, $U$ open and $y \in U$ implies $x \in {}^*U \subset {}^*A$).</p> <p>So let $\mathcal{X}$ be a set of topological spaces indexed by $I$, and $P$, the product of these spaces; write ${}^*P$ for the "non-standard product" of the set ${}^*\mathcal{X}$ of topological spaces indexed by ${}^*I$, and let $x \in {}^*P$. It suffices to show that $x$ is near-standard.</p> <p>For each $\kappa \in I$, let $x_\kappa \in {}^*X_\kappa \in {}^*\mathcal{X}$ be the $\kappa$th factor of $x$. Then $x_\kappa$ is necessarily near-standard, because $X_\kappa \in \mathcal{X}$ is compact. But this means we can find a point $y \in P$ with factors $y_\kappa \in X_\kappa$ such that $U \subset X_\kappa$ open and $y_\kappa \in U$ implies $x_\kappa \in {}^*U \subset {}^*X_\kappa$, thus $V \subset P$ open and $y \in V$ implies $x \in {}^*V \subset {}^*P$. But this means $x$ is near-standard, so $P$ is compact.</p> <p>"Under the hood," this is basically the ultrafilter proof (my favorite, to answer the original question), so the axiom of choice is required in more or less the same places: while the non-standard objects exist by the Boolean prime ideal theorem, "finding" the $y_\kappa$ in non-Hausdorff spaces requires the full axiom of choice.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/109458#109458 Answer by M Mueger for What is your favorite proof of Tychonoff's Theorem? M Mueger 2012-10-12T13:40:53Z 2012-10-12T13:40:53Z <p>I'm surprised that nobody has mentioned the proof using universal nets. (It can be found, e.g., in Pedersen's 'Analysis NOW' and in Bredon's 'Topology and geometry'.)</p> <p>A universal net in a set X is a net which, for every $Y\subset X$, ultimately lives in $Y$ or $X\backslash Y$. One easily sees that composition of a universal net in X with a function $f:X\rightarrow Y$ gives a universal net in $Y$. Using the ultrafiler lemma, one proves that every net has a universal subnet. All this involves no topology.</p> <p>Combining the above with standard facts, the proof of Tychonov is extremely short. All one needs is: - a space is compact if and only if every net has a limit point (equiv., a convergent subnet), - a net in $\prod_iX_i$ converges if and only if it converges coordinate-wise.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/112760#112760 Answer by M Mueger for What is your favorite proof of Tychonoff's Theorem? M Mueger 2012-11-18T14:15:18Z 2012-11-18T14:15:18Z <p>I have been teaching general topology for several years, but remained unsatisfied by the proofs given in the books that I based the course upon. Finally I wound up writing my own lecture notes, still not quite finished. In those notes, I give four different proofs. Two of them use (ultra)filters, but one of them avoids the terminology. The other two proofs use nets, namely Chernoff's proof without and Kelley's with universal nets.</p> <p>The notes can be found at <a href="http://www.math.ru.nl/~mueger/topology2012.pdf" rel="nofollow">http://www.math.ru.nl/~mueger/topology2012.pdf</a></p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/113821#113821 Answer by Joseph Van Name for What is your favorite proof of Tychonoff's Theorem? Joseph Van Name 2012-11-19T12:04:33Z 2012-11-19T12:04:33Z <p>Since all of the answers to this question(except the one involving Alexander's subbase lemma) refer to a usually strange rehashing of the ultrafilter proof (BOO), I decided to give two nice proofs to Tychonoff's theorem here for Hausdorff spaces.</p> <p>The first proof of Tychonoff's theorem for Hausdorff spaces uses the Stone-Cech compactification. This proof is useful when one constructs the Stone-Cech compactification before Tychonoff's theorem.</p> <p>Proof: Assume that <code>$X_{i}$</code> is compact for $i\in I$. Let <code>$X=\prod_{i\in I}X_{i}$</code> be the product space. Then each projection <code>$\pi_{i}:X\rightarrow X_{i}$</code> extends to a continuous map <code>$\overline{\pi_{i}}:\beta X\rightarrow X_{i}$</code> since each <code>$X_{i}$</code> is compact. Therefore the map <code>$f:\beta X\rightarrow X$ where $f(x_{i})_{i\in I}=(\overline{\pi_{i}}(x))_{i\in I}$</code> is a continuous surjection, so $X$ is compact being the continuous surjective image of $\beta X$. QED</p> <p>For the second proof we use the following facts about uniform spaces that every mathematician should be aware of.</p> <p>i. Every compact Hausdorff space has a unique compatible uniformity and that uniformity is complete and totally bounded.</p> <p>ii. If a uniform space is complete and totally bounded, then it is compact.</p> <p>Tychonoff's theorem then immediately follows from the fact that the product of complete uniform spaces is complete and that the product of totally bounded uniform spaces is totally bounded. And this proof is intuitive because it is easier to imagine that the product of complete and totally bounded uniform spaces is complete and totally bounded than to imagine that the product of compact spaces is compact.</p>