Talk:König's theorem (set theory)

Remark on Easton's theorem
The article has a remark that $$\kappa < 2^{\kappa}$$ 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.  &mdash; 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)

Unnamed Inequality
Does the theorem mentioned in passing:
 * IF $$m_i < n_i \!$$ for all i in I, THEN we can only conclude $$\sum_{i\in I} m_i \le \sum_{i\in I} n_i $$.

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)

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). &mdash; 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)
 * I don't think that's true, if I'm understanding you. If each $$A_i$$ is well ordered, then $$f(i) = \min A_i$$ is an element of the product. 128.237.226.38 (talk) 18:24, 18 August 2013 (UTC)
 * True, but a well-ordered cardinal corresponds to a well-orderable set. If each $$A_i$$ is well-ordered by $$<_i$$, then $$f(i)=\min^{<_i} A_i$$ would be well-defined.  But, if each $$A_i$$ is well-orderable, we still need the axiom of choice to select a well-ordering.  — Arthur Rubin  (talk) 18:46, 18 August 2013 (UTC)
 * Consequences of the Axiom of Choice notes, among other consequences, that the axiom of choice for a well-ordered family of well-orderable sets is not a consequence of ZF or NBG set theories. — Arthur Rubin  (talk) 18:46, 18 August 2013 (UTC)

Use "Kőnig" instead of "König" throughout?
Copied from here; please respond there.

Since this theorem was by a Hungarian named Kőnig, shouldn't the article properly use the double acute accent instead of the umlaut on his name? Just checking here. If nobody objects, I think I'll make the change. Oliphaunt (talk) 13:41, 9 May 2011 (UTC)


 * Since our biography of Gyula Kőnig says that he used the umlaut in the pseudonym, "Julius König", which he used when publishing mathematics, I think we should leave it as an umlaut. That is the name he chose to be called by when writing mathematics, and this article is about his mathematics. JRSpriggs (talk) 11:19, 10 May 2011 (UTC)

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.) &mdash; 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)

If the j in the proof is a projection onto B_j, we should probably say so. It took me a while to come to that conclusion, and that's time we can save people like me.

173.25.54.191 (talk) 23:26, 13 March 2014 (UTC)

Equivalent of < under AC
simplified the proof in König's theorem (set theory). However, he made a mistake in his handling of the empty set, which I fixed. Notice that the axiom of choice implies that all of the following are equivalent to each other: R.e.b. had just used "there is no surjection from A to B". This is not an equivalent because it is satisfied when A is non-empty and B is empty. I chose to replace it with the third option above, but I could have used the fourth. Is there any feeling among you-all that the fourth option would be better?
 * 1) A<B, i.e. there is an injection from A to B but no injection from B to A;
 * 2) there is no injection from B to A;
 * 3) there is no surjection from A to B and B is not the empty set; and
 * 4) there is no surjective partial function from A to B.

Is there any feeling that the equivalence chosen should be proved as a lemma? This would require proving that the axiom of choice implies that either A can be injected into B or B can be injected into A. This proof is rather difficult. JRSpriggs (talk) 07:59, 20 June 2014 (UTC)