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>,  <idscoord@sw.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 <elebeau@ens-lyon.fr> 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]

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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)