Newsgroups: sci.math Subject: Re: Tychonoff's Theorem From: rusin@washington.math.niu.edu (Dave Rusin) Date: 27 Jan 1995 22:40:50 GMT In article <3gbpuj$n3q@transfer.stratus.com>, wrote: >Quick Question (I don't have my topology book in front of me): > >What does Tychonoff's Theorem state? That the product of compact spaces is compact. This is difficult but interesting when you take an _arbitrary_ (not finite) product of compact sets, since compactness ought to have something to do with finiteness. (The crux of the matter is that it works precisely because the product topology is defined the way it is and not defined as what is sometimes called the "box" topology. Using the box topology the product of compact sets need not be compact). The result is useful when doing things like embedding a space into an uncountable product of copies of the interval [0,1]. [Imminent death of the net predicted again, if indeed people _already_ have the book and just don't want to go get it.] dave (who also abuses the net) ============================================================================== From: elebeau@ens-lyon.fr (Edouard Lebeau) Newsgroups: sci.math Subject: Re: Tychonoff's Theorem Date: 28 Jan 1995 15:16:17 GMT The proof of Tychonoff's theorem for a finite product of compact spaces is easy. There is also a somewhat `natural' proof for a countable product of compact metric spaces, using Cantor's diagonal arguments. The general proof of Tychonoff's theorem can be found in lots of topology book. The most common proof uses filters and ultrafilters. For those who are unfamiliar with filters, a `more topological' proof can be found in W. Rudin's `Functional Analysis'. It uses both Zorn's lemma and Hausdorff's Maximality Theorem (which are both equivalent to the Axiom of Choice). > The result is useful when doing things like embedding a space into an > uncountable product of copies of the interval [0,1]. Another classical use of Tychonoff'Theorem in functional analysis is Banach-Alaoglu's Theorem: In the dual space of a Banach space, the unit ball is compact for its weak topology. This result often makes up for the fact that the unit ball is not compact for its metric topology, when the space is infinite-dimensional. [signature deleted - djr] ============================================================================== From: magidin@jaffna.berkeley.edu (Arturo Magidin) Newsgroups: sci.math Subject: Re: Tychonoff's Theorem Date: 28 Jan 1995 20:50:36 GMT In article <3gdn41$fg5@cri.ens-lyon.fr>, Edouard Lebeau wrote: [The proof of Tychonoff's Theorem...] >It uses both Zorn's lemma and >Hausdorff's Maximality Theorem >(which are both equivalent to >the Axiom of Choice). In fact, so is Tychonoff's Theorem, as it can be used to prove the Axiom of Choice; the result is due to (I believe) Kelley. [signature deleted - djr] ====================================================================== From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener) Newsgroups: sci.math Subject: Re: Tychonoff's Theorem Date: 29 Jan 1995 14:16:38 GMT In article <3geams$el7@agate.berkeley.edu>, magidin@jaffna (Arturo Magidin) writes: >In fact, so is Tychonoff's Theorem, as it can be used to prove the >Axiom of Choice; the result is due to (I believe) Kelley. Note that Tychonoff's theorem for T1 spaces (points are closed) is equivalent to AC. For the more common application involving Hausdorff spaces, the theorem is equivalent to "the Prime Ideal Theorem", which says, in one form, that every non-trivial filter in a Boolean algebra can be extended to an ultrafilter. The reason for the difference is that Hausdorff spaces already have a bit of free choice in them--filters and nets have _unique_ limits. -- -Matthew P Wiener (weemba@sagi.wistar.upenn.edu)