Talk:Tychonoff's Theorem

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Various questions

a) The definition for "tree":

In the definition for "tree", the symbol for the ordering given is for a generic strict ordering, while the definition for a partially ordered set as given on ProofWiki calls for a standard partial ordering which includes reflexivity.
Similarly, there is a mention of "strictly bounded from above". I have never seen the term "strictly bounded" before and I have had difficulty finding a reliable citation for its definition.

Question is: how important is the "strictly bounded" and the "strict ordering" in the above definition?

b) There's more things, but I have to go and do something else now, I will be back. --prime mover 03:16, 31 December 2011 (CST)

I assume that there are two equally good definitions of a partially ordered set --- with or without strict inequality. The definitions of a well-order relation and of a tree in Wikipedia use strict order. Since one is expressible in terms of the other, i do not think this is an issue, maybe a note needs to be added to the definition of poset in ProofWiki.
It's a good point, but as you say not important (and the fact that it is not important is also a good point!) so the definitions in the pages are to be amended so as to allow for such ordered sets to be defined using strict orderings. Another page may then be added to explain that the two styles of definition are equivalent.--prime mover 07:29, 31 December 2011 (CST)
I put "strict" in parenthesis to show that it can be omitted from the sentence without change of meaning. I put it because "open" intervals are often used in the proof. You are right that these definitions need to be removed from the article and it needs to be linked to other articles. --Cokaban 04:57, 31 December 2011 (CST)
This last exercise is under way. --prime mover 07:29, 31 December 2011 (CST)


To have equations flushed left, it would be easier to set an option like \documentclass[fleqn], and then use \[ \], is this possible? --Cokaban 05:17, 31 December 2011 (CST)

Not as far as I know. This is not a full $\LaTeX$ environment. Instead it uses $\LaTeX$ structures within a MediaWiki environment, and as such uses MediaWiki page formatting.
In order to get equations to flush left, use this technique:
$d := b^2 - 4 a c$
Simple as that. Colons are used to indent.
We do have a technique for aligning equations, which is used on most pages on this site. Click "Random proof" and it will (eventually) take you to a page where you will see it used.
I plan to work through the page and convert it to house style (e.g. each definition and lemma on its own page, dollar-signs not backslash-bracket for delimiters, equations are deliberately not centered on the page, sentences are kept short, and one sentence per line, and so on), but it takes time to do. --prime mover 06:44, 31 December 2011 (CST)

Functions ordered by inclusion and intervals

By "initial interval" in a well-ordered set i mean any subset which with every element contains all smaller elements. In general posets there could be used the term "convex subset", but for well-ordered sets "interval" shouldn't be ambiguous.

A function f is included in a function g if f is a subset of g. --Cokaban 20:17, 31 December 2011 (CST)


Your comment: "As the Axiom of Choice is not assumed, its consequences should be put in parenthesis or removed from the statement." If it's not assumed, and is therefore not relevant for the purposes of the proof, then it needs to be removed completely. If it is assumed, and in fact is needed (to justify the fact that $(I,<)$ is well-orderable) then it does need to be made part of the proof.

As it is, the proof itself appears to prove Tychonoff's Theorem only in the case where the indexing set is well-orderable. In order to prove it for every set, one needs to make the step to "the indexing set is guaranteed to be well-orderable by the Well-Ordering Theorem" and we're back to needing AoC again. --prime mover 01:54, 1 January 2012 (CST)

Exactly, this is why the two statement that i've put in parentheses were initially footnotes. I think they should be footnotes. --Cokaban 03:43, 1 January 2012 (CST)

Link to a proof that the product of two compact spaces is compact.

I do not know if a link to a proof that the product of two compact spaces is compact is indeed needed. The reverence i made was about the technique, not about the result, so if we make a link, and then somebody will change the proof for two space, the link will make no sense. I prefer it to be just a vague remark in parentheses to help understand the connection with the finite case. --Cokaban 04:07, 1 January 2012 (CST)

That's not how this site works. Every single statement needs to be backed up by a proof. Therefore it will need to be done, which is why I added the placeholder. We pride ourselves on precision, and try not to rely on "just a vague remark in parentheses". --prime mover 05:26, 1 January 2012 (CST)
This remark is not a part of a proof, it does not need to be backed up. It is intended to help follow the proof, like a picture would. You can remove it without loss of precision. --Cokaban 07:42, 1 January 2012 (CST)
Better to include that proof for two space. If "somebody will change the proof for two space, the link will make no sense" applies to every single proof on this site. With that philosophy it would be doubtful if anything were ever linked to anything. --prime mover 08:21, 1 January 2012 (CST)
I mean, if you cite and link results, then changing proofs does not break links, but here i cite a particular common technique of a proof, which should be familiar for a big part of readers, but does not have to be the only one used on this site. --Cokaban 08:30, 1 January 2012 (CST)

Applications section

"The proof that $\left[{0 .. 1}\right]^\Z$ is compact does not require the Axiom of Choice."

Yes, I see that written down, and I see the supposed explanation, but what is missing is exactly why. Therefore the placeholder "explain" tag was put in place for a reminder for this to be done. Please do not remove templates from pages until the issue has been resolved.

Besides, it is a completely separate result and needs separate page for it, to which we may link from here. --prime mover 05:26, 1 January 2012 (CST)

The tag is not very clear to me, it seems to ask to explain why the proof that the infimum of a closed subset of [0..1] belongs to this closed subset can be made without AoC, but in my opinion there is nothing to explain here. The tag needs to say more clearly what needs to be explained, or just ask for details, but there is nothing to explain in why the infimum of a closed set is in it regardless of the AoC. --Cokaban 08:18, 1 January 2012 (CST)
There is a statement being made here "the product of all nonempty closed subsets of $\left[{0 .. 1}\right]$ contains the greatest lower bound function $\inf$" which needs to be demonstrated or expanded. That's all I mean. Also, that statement needs to go on a separate page. --prime mover 08:24, 1 January 2012 (CST)

Greatest vs. maximal

Again, an "explain" template has been removed without any explanation having been done. In order for this page to be understandable, the difference between "greatest" and "maximal" needs to be explained. Perhaps this is the result of unfamiliar terminology for well-understood concepts, in which case a link needs to be added to exactly which concepts these are. If not, then a page needs to be added for "greatest element" and an explanation needs to be made as to why they are different. --prime mover 05:31, 1 January 2012 (CST)

The difference between "greatest" and "maximal" does not need to be explained on this page, it is enough to link to their definitions, which are the standard ones. --Cokaban 07:45, 1 January 2012 (CST)

A page for "greatest element" is to be added to this site. The difference between "greatest" and "maximal" is not well explained in the literature. --prime mover 08:21, 1 January 2012 (CST)

Another instance of AoC

"according to the hypotheses, $T_{\mathcal O}$ has a branch $B$."

My point was that the hypotheses contain that supposition. If in fact it does not have a branch, then the proof does not hold for such cases.

It is assumed in the statement that in fact it does have a branch. --Cokaban 07:52, 1 January 2012 (CST)
I mean, every such "special" subtree of F. --Cokaban 08:05, 1 January 2012 (CST)

What I am trying to communicate is that without the AoC, the proof applies only to where the indexing set has a well-ordering.

This is the assumptions (hypotheses) of the theorem. The proof of any theorem only applies to its hypotheses. --Cokaban 07:52, 1 January 2012 (CST)

If AoC does not hold, then there are sets which can not be ordered. Therefore there are trees which do not have branches. Then Tychonoff's Theorem can not be proved for such sets. So your AoC-less proof of this theorem only applies to indexing sets which can be well-ordered and therefore can not be used for the general case unless AoC is taken to hold. We're back where we started, a proof which relies upon AoC. --prime mover 05:38, 1 January 2012 (CST)

Exactly, this is why that proof does not need the Axiom of Choice. With the axiom of Choice, the usual statement can be easily deduced form it, using Well-Ordering Principle and Hausdorff Maximal Principle. --Cokaban 07:52, 1 January 2012 (CST)
The usual statement of TT that assumes the AoC does not allow to deduce easily that the product of two compact spaces is compact without AoC, and that [0..1]^Z is compact without AoC. --Cokaban 08:02, 1 January 2012 (CST)
Sorry I'm still not getting it. The fact that the AoC is assumed as one of the hypotheses means that this proof of Tychonoff's Theorem depends on the AoC. So calling it "Theorem without Axiom of Choice" is misleading, as it implies "Here is a proof of TT which does not depend on the AoC." But it does. One of the hypotheses is AoC. --prime mover 08:09, 1 January 2012 (CST)
AoC is not assumed. I agree that this theorem is different from the usual one, this is why i gave a second statement. The usual theorem does not hold without AoC, but this one does. Do you have a better name for it than "Theorem without AoC"? --Cokaban 08:25, 1 January 2012 (CST)

Greatest element of what

"A set S has a maximal element" means it has a maximal element of itself.

"A set S has the greatest element" means there is the greatest element in it (i am not sure if from the point of view of correct English it should be "a greatest" or "the greatest").

"A function f has a maximum" means it attains its maximum. --Cokaban 07:58, 1 January 2012 (CST)

I understand now. This site is missing a page for "greatest element" (and similarly, for "least element"). I will attend to this in due course. --prime mover 08:21, 1 January 2012 (CST)

Merging with the finite product case

I am against merging in the Topological Product of Compact Spaces, that other fact has a sufficiently simple proof without the Axiom of Choice, and uses a slightly different language (ordered pairs instead of functions or sequences). --Cokaban 05:03, 2 January 2012 (CST)

Yes, I think I agree. The plan is now to gather all the separate proofs of this result, put them into their own pages, and transclude them into one master page. But not yet, I am doing something else at the moment. --prime mover 16:36, 2 January 2012 (CST)

If and only if

Can we please dispense with the "only if" here? That follows directly from the fact that projections are continuous and continuous images of compact sets are compact, so it feels kind of like an "oh, and by the way" add-on. --Dfeuer (talk) 08:32, 2 January 2013 (UTC)

If it turns out that T's T does not itself include the "only if" part, as I suspect might be the case, then I'm prepared to agree with you. However, we need to check that there are no links to this page which do invoke the only-if part (which, trivial though it is, itself needs to be documented). Those will need to be amended so as to point to this new page: compact product implies compact factors. --prime mover (talk) 13:28, 2 January 2013 (UTC)
I checked the sources I have available: Kelley and Munkres don't include the "only if". S & S do include it. --Dfeuer (talk) 18:39, 2 January 2013 (UTC)


One option is to use a lemma that imposes the condition that filters on the product converge if and only if their projections do. That would push the meat of the proofs of the general case and the case for Hausdorff spaces into that lemma. --Dfeuer (talk) 18:44, 2 January 2013 (UTC)