Munkres Chapter 5: Problem involving the Tychonoff Theorem

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Summary:
Tychonoff Theorem
Hi,
In Chapter 5 Munkres proves the Tychonoff Theorem and after proving the theorem the first exercise is: Let ##X## be a space. Let ##\mathcal{D}## be a collection of subsets of ##X## that is maximal with respect to finite intersection property
(a) Show that ##x\in\overline{D}## for every ##D\in\mathcal{D}## if and only if every neighborhood of ##x## belongs to ##\mathcal{D}##.
(b) Let ##D\in\mathcal{D}##. Show that if ##A\supset D##, the ##A\in\mathcal{D}##.
(c) Show that if ##X## satisfies the ##T_1## axiom, there is at most one point belonging to ##\bigcap_{D\in\mathcal{D}}\overline{D}##

(a) and (b) are trivial consequences of the lemma 37.2 proven before the Tychonoff Theorem's proof. On the other hand, (c) would be of the same level of triviality if ##T_2## was assumed, however, as written only ##T_1## is assumed.

Could it be just a Typo? or the problem can be solved assuming only ##T_1##. Any comments or help will be appreciated.
 
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  • #2
Office_Shredder
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I think you are right. A simple counterexample, let ##X=\mathbb{Z}## with the cofinite topology. Then let ##\mathcal{D}## be the set of all infinite subsets. Every finite intersection is nonempty, and if you include any finite set then you can get a trivial intersection of that set with its complement, so this is a maximal collection. The closure of every infinite set is ##X##, so
$$\bigcap_{D\in \mathcal{D}} \overline{D}=X$$
Which contradicts the claim.
 
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  • #3
mathwonk
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This question reminds me that I have, as a practicing mathematician for over 40 years, never, ever, needed the Tychonoff theorem, at least not in the case of infinitely many factors. so to me, the interesting case is a finite product. even a normal human might come up with a proof in that case. just a remark, from a person rather ignorant of tychonoff/logic etc... but you might try it for a factors.
 
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I think you are right. A simple counterexample, let ##X=\mathbb{Z}## with the cofinite topology. Then let ##\mathcal{D}## be the set of all infinite subsets. Every finite intersection is nonempty, and if you include any finite set then you can get a trivial intersection of that set with its complement, so this is a maximal collection. The closure of every infinite set is ##X##, so
$$\bigcap_{D\in \mathcal{D}} \overline{D}=X$$
Which contradicts the claim.
Excellent! However, I think it needs a slight modification. I believe we should take ##\mathcal{D}=\mathcal{F}##, i.e., take the maximal set with respect to the property of being open and with the finite intersection property. Not every two infinite sets have not an empty intersection. Thank you very much.
 
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This question reminds me that I have, as a practicing mathematician for over 40 years, never, ever, needed the Tychonoff theorem, at least not in the case of infinitely many factors. so to me, the interesting case is a finite product. even a normal human might come up with a proof in that case. just a remark, from a person rather ignorant of tychonoff/logic etc... but you might try it for a factors.
I suppose Tychonoff's usefulness might depend on the fields of mathematics one is working in.
 
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Office_Shredder
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Excellent! However, I think it needs a slight modification. I believe we should take ##\mathcal{D}=\mathcal{F}##, i.e., take the maximal set with respect to the property of being open and with the finite intersection property. Not every two infinite sets have not an empty intersection. Thank you very much.

Agh, yes, I meant all the infinite sets in the topology. Obviously the other ones don't really exist 😆
 
  • #7
mathwonk
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@facenian: you got me thinking and wondering who uses it. It was taught to me by an abstract analyst (Lynn Loomis), and we did use it in that course to prove certain abstract embedding theorems. (You find a lot of bounded real valued functions on your space, and then regard those as coordinate functions on the product space of all their target intervals, sending each point of your space to the collection of all values at that point of all functions. I guess this is the starting point for classifying compactifications of a given completely regular space, which is pretty, i.e. choosing different collections of functions gives different embeddings, hence different compactifications. This kind of abstract stuff is not if much interest in the kind of more concrete and classical geiometry I do.) But then I found a paper reminding me it is equivalent to the axiom of choice, which I use all the time. So I guess to me it is an exotic, much less useful, version of AxCh! Thanks for the reminder that everything is interesting, to someone anyway.
 
  • #8
This question reminds me that I have, as a practicing mathematician for over 40 years, never, ever, needed the Tychonoff theorem, at least not in the case of infinitely many factors. so to me, the interesting case is a finite product. even a normal human might come up with a proof in that case. just a remark, from a person rather ignorant of tychonoff/logic etc... but you might try it for a factors.
You can use it to give an easy proof of the Banach-Alaoglu theorem, which is absolutely fundamental in abstract functional analysis. And as you said, it can also be used to show that certain compactifications, such as the Stone–Čech compactification, exist.
 

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