6
$\begingroup$

In the midst of learning about compactness I come across Tychonoff's Theorem:

Let $\{X_i : i \in \mathcal{A}\}$ be any collection of compact spaces. Then $\displaystyle\prod_{i \in \mathcal{A}}X_i$ is compact in the product topology.

I've just come from the fact that a finite product of compact spaces is compact, and I also know from studying bases of topologies that uncountable products aren't necessarily as nice (for example, the box topology has some problems for uncountable products).

The proof for Tychonoff's Theorem is:

Omitted (this is much harder than anything we have done here).

Internet searches lead to math overflow and topics that are very outside of my comfort zone.

Is there a proof of Tychonoff's Theorem for an undergrad?

$\endgroup$
5
  • 5
    $\begingroup$ Lots of undergrads learn the proof in Munkres' book "Topology". $\endgroup$
    – Lee Mosher
    May 27, 2016 at 17:44
  • $\begingroup$ @LeeMosher Thanks for that, it turns out that that book is also the recommended text for the course. $\endgroup$ May 27, 2016 at 17:53
  • $\begingroup$ The proof in General Topology by R. Engelking is by elementary means, is fairly short, and covers all cases (finite,countable,uncountable products) at once. $\endgroup$ May 27, 2016 at 17:56
  • $\begingroup$ @user254665 Thanks for the reference! There's a Tychonoff's theorem for finite products? $\endgroup$ May 27, 2016 at 18:01
  • $\begingroup$ There are ways to show that a product of finitely many compact spaces is compact, that don't generalize to infinite products. But it's more efficient to have a method that works in all cases. $\endgroup$ May 27, 2016 at 21:34

1 Answer 1

7
$\begingroup$

Perhaps the most elementary proof is the one that I first encountered as a freshman, using the Alexander subbase lemma. It requires Zorn’s lemma, but it does not require knowledge of filters, ultrafilters or nets. It’s carried out completely in this PDF. (And Alexander’s result is of some interest in its own right.)

$\endgroup$
9
  • 1
    $\begingroup$ Thanks for the really accessible proof - I've come across Zorn's lemma a couple of times, but I guess it's time to learn it properly now. $\endgroup$ May 27, 2016 at 17:54
  • 1
    $\begingroup$ @IrregularUser: You’re welcome. Yes, Zorn’s lemma is definitely something to learn; you’re especially likely to want it in abstract algebra courses. $\endgroup$ May 27, 2016 at 17:56
  • $\begingroup$ To the OP : There is a huge list of theorems that have been shown to be equivalent to, or to require Zorn's Lemma . It's like the number $\pi.$ You just cannot get away from it. $\endgroup$ May 27, 2016 at 17:59
  • $\begingroup$ @user254665 It seems so... much like the axiom of choice I guess. A year ago when texts would comment on the axiom choice it would be something skippable, but now it's being mentioned often enough that I felt that it was better to learn it than ignore it. $\endgroup$ May 27, 2016 at 18:03
  • $\begingroup$ Zorn's Lemma is equivalent to the axiom of choice (AC) in the axiom system ZF (Zermelo-Fraenkel). ... ZFC (which is ZF+AC) is the most popular axiomatic foundation system. $\endgroup$ May 27, 2016 at 21:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .