Talk:Kőnig's lemma

message: "It is known that there are finitely branching computable subtrees of w^<w that have no arithmetical path" --  I believe this is false and needs amending.
Suppose T is computable and finitely branching.  Then with a Sigma-1 oracle we can answer queries of form, "does node n have a successor node of index > m?"; using this oracle we can enumerate all successors of any given node n and recognize when there are no more.  So our tree's branching is Sigma-1-computably bounded, and so by the Sigma-1-relativized version of the result "Any computably-bounded-branching recursive tree has a Sigma-1-computable infinite path", our tree has a Sigma-1^{Sigma-1} = Sigma-2 -computable path.
I believe the correct result is  "there are computable subtrees (not locally finite) of w^<w that have infinite paths but no arithmetical infinite path".  But I don't have a reference.
----Andy Drucker  —Preceding informally signed comment added by 66.117.149.88 (talk) 04:37, 23 December 2006
:Yes, this was a typo, and your argument is sound. Please feel free to fix these things if you find them. I added a reference to one canonical source for a computable tree with paths but no hyperarithmetical ones. — Carl (CBM · talk) 23:29, 2 February 2009 (UTC)
thread_title:  Claim about hyperarithmetical paths 

message: The proof shows that for any vertex v1, and for any integer n, there exists a simple path of length n starting at v1.  The proof then claims that by induction an infinite path exists.  This is FALSE, induction on n shows that a claim is true for arbitrarily large finite values, but does not prove a anything about the infinite case.  
The proof thus glosses over the most essential part of the proof of Konig's lemma, showing that arbitrarily long finite simple paths ==> an infinite simple path.   Please explain if my analysis is mistaken.  Otherwise the Wikipedia proof should be corrected.  Thanks  —Preceding unsigned comment added by 128.135.199.198 (talk) 07:15, 23 November 2008
The fact the the trees produced by induction form a chain under the partial order of inclusion can possibly be used in the proof.—Preceding unsigned comment added by 128.135.199.198 (talk) 07:18, 23 November 2008
The proof is fine.  It shows that any v_i with infinitely many descendants has a child v_i+1 (not equal to any v_j for j <= i) with infinitely many descendants.  We assume v_0 has infinitely many descendants.  So induction gives a sequence {v_i} that is an infinite path by construction.  "This" is FALSE in general of course, but not in the case where each vertex has finitely many neighbors.  Previous comment is setting up a standard Zorn type argument.  — Preceding unsigned comment added by 173.18.240.48 (talk) 12:10, 5 October 2012 (UTC)
replies: text: No, the proof is not fine. The note after it about the Axiom of Choice, and the discussion below, explain why it is not fine. It needs to be modified to use the restricted form of dependent choice or the axiom of countable choice for finite sets. --Dfeuer (talk) 18:50, 27 May 2013 (UTC)
thread_title: Potential problem with the proof of Konig's lemma presented on this page

message: This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 04:14, 10 November 2007 (UTC)
thread_title:  WikiProject class rating

message: It would be useful to have a separate page for Brouwer's fan theorem. Tkuvho (talk) 13:18, 20 September 2010 (UTC)
thread_title:  fan theorem 

message: Copied/edited from here; please respond there.
Since this lemma 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)
replies: text: Are there may English language sources that use the alternate diacritic? In the ones I am familiar with, texts in mathematical logic, they use the umlaut mark. — Carl (CBM · talk) 13:46, 9 May 2011 (UTC)

text: The linked-to discussion no longer really covers this case as, referring to the father of the person in question here, it has been resolved with reference to the fact that that author chose to identify himself using the umlaut in mathematical publications. Unless a similar fact is known with reference to the author in question here, that discussion is no longer useful. I personally am presently on the side of spelling the author's name correctly (i.e. with double acute accent) unless an alternative preference on the author's part becomes known. The fact that sources are too lazy to spell an author's name correctly doesn't seem sufficient to me to warrant asserting the name of a lemma named for a specific individual should have a name different to that of the individual concerned, which would strike me as quite a bizarre precedent. 130.88.174.147 (talk) 19:19, 30 May 2011 (UTC)
thread_title:  Use "Kőnig" instead of "König" throughout?