Tychonoff theorem and Zorn's lemma

[Fernsanz]

Hi.

I'm trying to understand why is the Zorn's lemma needed to prove the Tychonoff theorem -the product of compact spaces is compact-.

Specifically my question is about the proof in the book from Bachman-Narici based on the technic of Finite Intersection Property that you can find here:
http://books.google.es/books?id=wCH...resnum=1&ved=0CBcQ6AEwAA#v=onepage&q&f=false"

Why can not we just carry over the proof of the finite case? I mean, once the point \hat{x} has been proposed as the common adherence point and bearing in mind the basis for the topology of X consists of the sets
\prod V_{\alpha} where V_{\alpha} is an open set V_{\alpha} \subset X_{\alpha} for finitely many \alpha and X_{\alpha} for the rest, it is obvious that any of these basis set containing \hat{x} will also intersect all of E^{\gamma}. So, where and why is the Zorn's lemma needed?

Thanks in advance.

 

[g_edgar]

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Presumably "choice" is used to define the point \hat{x}

 

[Fernsanz]

Too many people have viewed that page, no more allowed...
Presumably "choice" is used to define the point \hat{x}

That would be more logical for me, but no proof mention the "choice" to define the point \hat{x} even though, in mi opinion, if one is to use the Axiom of Choice/Zorn lemma this is the step to mention it. By contrast, all proofs use the AC/Zorn after the point have been selected to deduce that it is a common adherence point; to me, this fact seems obvious by the very definition of the product topology base without relying on any other argument.

 

[Fredrik]

Too many people have viewed that page, no more allowed...

You can view the page if you just change the country code in the URL. I changed .es (Spain) to .se (Sweden) and it worked fine.

 

[Fernsanz]

No answers?

 

[Fredrik]

Nope, but by coincidence I'm studying the proof of Tychonoff's theorem right now, so I might have something profound to say tomorrow. What I can say right now is that I think you would like the discussion preceding the proof at the start of chapter 5 of Munkres.

Edit: Here's a quote from the bottom of page 230 (the first page of chapter 5), regarding the approach that's used in the finite case:

It's quite tricky to make this approach work for an arbitrary product of compact spaces; one must well-order the index set and use transfinite induction. (See exercise 5). An alternate approach is [...] using Zorn's lemma.​

So it seems that it is possible to do it in a way that's similar to the proof of the finite case, without using Zorn, but using the well-ordering theorem and transfinite induction instead. I'm thinking that the proof based on Zorn is probably more popular because more people are familiar with Zorn's lemma.

 

[Fredrik]

I understand the proof in Munkres now. I only had a very quick look at the proof in Bachman & Narici. It appears to be the same. If you want to understand why we need to make the family of sets with the finite intersection property maximal (the existence of the maximal family is what Zorn is used for), the discussion preceding the theorem in Munkres will definitely help. It includes an example of how this method of proof can fail when we try to use a family that isn't maximal.