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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?

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    $\begingroup$ Lots of undergrads learn the proof in Munkres' book "Topology". $\endgroup$ – Lee Mosher May 27 '16 at 17:44
  • $\begingroup$ @LeeMosher Thanks for that, it turns out that that book is also the recommended text for the course. $\endgroup$ – Irregular User May 27 '16 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. $\endgroup$ – DanielWainfleet May 27 '16 at 17:56
  • $\begingroup$ @user254665 Thanks for the reference! There's a Tychonoff's theorem for finite products? $\endgroup$ – Irregular User May 27 '16 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. $\endgroup$ – DanielWainfleet May 27 '16 at 21:34
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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.)

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    $\begingroup$ 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. $\endgroup$ – Irregular User May 27 '16 at 17:54
  • $\begingroup$ @IrregularUser: You’re welcome. Yes, Zorn’s lemma is definitely something to learn; you’re especially likely to want it in abstract algebra courses. $\endgroup$ – Brian M. Scott May 27 '16 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. $\endgroup$ – DanielWainfleet May 27 '16 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. $\endgroup$ – Irregular User May 27 '16 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. $\endgroup$ – DanielWainfleet May 27 '16 at 21:30

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