Talk:König's theorem (set theory)

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Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Geometry guy 22:55, 10 July 2007 (UTC)

Contents 1 The proof of König's theorem 2 Remark on Easton's theorem 3 Unnamed Inequality 4 Connections to the AC

[edit] The proof of König's theorem

Although my modifications to the proof may appear to be WP:OR, I was just adjusting User:JRSpriggs's proof to have a consistent notation. (There is also a proof in Equivalents of the Axiom of Choice, as well as in PlanetMath.) — Arthur Rubin | (talk) 22:16, 8 July 2006 (UTC)

Thanks for your help. I was going to edit it some more, but you saved me the effort. JRSpriggs 06:58, 9 July 2006 (UTC)

Well, some of the effort. I found some more changes that needed to be made. JRSpriggs 09:52, 9 July 2006 (UTC)

[edit] Remark on Easton's theorem

The article has a remark that κ < 2^κ is the only restraint on the continuum function according to Easton's theorem. This isn't literally true; monotonicity at least is also required. Also, I am used to seeing Easton's theorem's restrictions stated in terms of cofinality rather than just cardinality. Am I missing something? CMummert 13:24, 29 August 2006 (UTC)

I don't think you're missing anything. The article on Easton's theorem seems correct, and has the other constraints. — Arthur Rubin | (talk) 13:46, 29 August 2006 (UTC)

I made a minimal edit to correct the problem, but I will not be upset if someone more attached to the page edits it further. CMummert 00:30, 30 August 2006 (UTC)

[edit] Unnamed Inequality

Does the theorem mentioned in passing:

IF for all i in I, THEN we can only conclude .

have a name or a reference? --Michael C. Price ^talk 03:36, 30 August 2006 (UTC)

This is too obvious to really call it a theorem. To see that it is less than or equal, you just combine the injections together into one big injection. To see that it cannot be strictly less in all cases, we just have to provide one counter-example. Let the index set be the natural numbers. Let m_i = 1 in every case. Let n_i = 2 in every case. Then both sums are aleph-null, hence equal. Use m_i maps to {i}; and n_i maps to {2i, 2i+1}. OK? JRSpriggs 05:51, 30 August 2006 (UTC)

Thanks, I took the liberty of inserting your explanation into the article for the benefit of maths dummies such as myself. --Michael C. Price ^talk 08:16, 30 August 2006 (UTC)

[edit] Connections to the AC

I think "However, this formulation cannot even be stated without the use of some form of the Axiom of Choice." is wrong if one uses the definition c is a cardinal, if it is an ordinal and if there is no bijection to a smaller ordinal. You can state König's theorem then and I think you can even prove it without AC. (by Gockel)

Hmmm. I think you're correct. That shows the deficiency of that definition of cardinal in this context (and in the absence of AC). — Arthur Rubin | (talk) 12:22, 8 October 2006 (UTC)

Surely you cannot prove it without the axiom of choice, as the article says that it implies the axiom of choice, and the axiom of choice is not provable. --PhiJ 21:59, 22 November 2006 (UTC)

What Gockel means is that you can prove König's Theorem for well-ordered cardinals without AC. I don't know if this is correct, but there's no obvious problem, as this weak form of König's theorem does not appear to imply AC. --Zundark 22:51, 22 November 2006 (UTC)

I'm not sure that this statement actually works; you need AC to show that an infinite product of well-ordered cardinals is well-defined, for example.209.210.225.6 19:23, 6 April 2007 (UTC)