Talk:Hausdorff Maximal Principle

The link to AoC has been removed. Can this be explained? --prime mover 01:28, 26 January 2012 (EST)
 * Nope. It is equivalent to Zorn's Lemma (or so I thought). Furthermore, it is used in the proof that 'According to Zorn's Lemma...' --Lord_Farin 02:26, 26 January 2012 (EST)
 * What I thought. I put it back in again. Admins always win edit wars. --prime mover 03:49, 26 January 2012 (EST)
 * I removed it -- not because AC isn't used in the proof, but because the AC entry is inaccurate in that it overstates the objections to AC and, thus, misrepresents the actual status of AC.--ratboy 05:51, 15 February 2012 (EST)
 * Okay, I understand. Your viewpoint in itself I see more as a philosophical position than a statement of truth. Some mathematicians are still genuinely in denial about AoC. Such is the position of this website. If you take the word "controversial" to mean "subject to ongoing controversy" (and in this context "controversy" means "some people think it's true, some people don't") then AoC is controversial. While it is legitimate to discuss whether this standpoint is valid or not, that point is unimportant compared with the undisputable fact that HMP relies upon AoC. Therefore, for all its perceived possible faults, the link itself should I believe remain. --prime mover 06:17, 15 February 2012 (EST)

I've removed the references to the other two proofs because I don't think they're clearer and because by not using Zorn's lemma you can use the Haussdorff maximal principle to provide a much nicer proof of Zorn's Lemma than the one currently present. Is this ok? --DRMacIver 11:41, 17 February 2012
 * See your talk page. As for the circularity problems, they might be dealt with later. It is not a problem to prove Zorn's Lemma using HMP because this will establish equivalence of these results (in a direct manner). This whole field may be up for restructuring some time, but that's not for today. --Lord_Farin 06:55, 17 February 2012 (EST)

Proof 1 is essentially Proof 2 with the details omitted: the greater part of Proof 2 is the verification of the claim "Since each set in c is itself a chain, and hence an element of S, the union of these chains is also a chain, and an element of S." Also Proof 2 proves the slightly stronger result that every chain in a poset P is included in a maximal chain in P, a fact which ought to be noted. --ratboy 05:00, 28 February 2012 (MST)


 * Indeed it ought. --prime mover 01:58, 29 February 2012 (EST)