Talk:Cantor's paradox
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[edit] Need more context
I've given this article a general cleanup, but I think it could use a little more:
- The formal statement of Cantor's paradox isn't so great.
- Some context explaining why Cantor's paradox is significant would be very useful here.
If we can fix those two things then I think the {{attention}} tag can be taken away. Tim Pierce 06:30, 12 December 2005 (UTC)
- I've done a major revision on the page and I think I've addressed your concerns, so I'm excising the {{attention}} tag. Ryan Reich 17:16, 21 January 2006 (UTC)
[edit] NBG set theory
"The fact that NBG set theory resolves the paradox is therefore a point in its favor as a suitable replacement."
What is NBG set theory? This should either be a link, or it should be removed. Surely axiomatic set theory is the standard 'replacement' for naive set theory anyway. InformationSpace 00:38, 4 May 2007 (UTC)
- Von Neumann–Bernays–Gödel set theory, but, as well-known, ZFC also solves the paradox in the same manner, and you can deal with classes in ZFC (but not as objects of the theory). The sentence has to be removed. 81.57.95.102 12:41, 8 May 2007 (UTC)
[edit] AC and 'Since the cardinal numbers are well-ordered' ?
The first sentence of Cantor's_paradox#Discussion and consequences depends on the axiom of choice, doesn't it? I'm no expert, could somebody make clear if this is dependent or independent on the axiom of choice? Aaron McDaid (talk - contribs) 07:40, 2 June 2007 (UTC)
- Well, it's a bit of a non-sequitur to talk about it being dependent on axioms (AC or otherwise), since the paradox was posed in a non-axiomatic context. But then the whole "Discussion and consequences" section is a bit of a non-sequitur, as far as that goes. In any case it's trivial to see that there is no largest ordinal, for the same reason there's no largest natural number (every ordinal has a successor). --Trovatore 07:45, 2 June 2007 (UTC)
- I must have mixed up the two sections 'Statement and proof' and 'Discussion and consequences'. However, I still think their may be gaps in the article with regard to the axiom of choice (AC). Is it true that cardinals can ordered with or without AC? And that AC is only required to well-order them? I'm no longer implying there's a flaw in the proof, just being curious with a view to improving the article if appropriate. Aaron McDaid (talk - contribs) 08:55, 2 June 2007 (UTC)
- The statement that the cardinalities are wellordered -- or, indeed, even linearly ordered -- is equivalent (using ZF as the background theory) to AC.
- As for improving the article -- not meaning offense to the author(s), I frankly think it could use a nearly complete rewrite. The paradox is that, if you accept the apparently reasonable proposition that it should be possible to collect all sets into one completed totality, then that object ought to have the maximum possible cardinality. But then what about the powerset of that object, which must have a greater cardinality?
- From the article as it stands, in my opinion, it's very hard to understand either what the problem is, or the resolution (which is simply that it's not possible to collect all sets into a completed whole). --Trovatore 09:17, 2 June 2007 (UTC)
- I had thought that the Cantor–Bernstein–Schroeder theorem tells us that the cardinals can be ordered without using AC. But I'm not so sure now. It says that if |A|<=|B| and |B|<=|A|, then |A|=|B|. Was I mistaken in assuming that it's always true that |A|<=|B| and/or |B|<=|A| ? i.e. Without AC there may be a pair of cardinalities that cannot be ordered? Aaron McDaid (talk - contribs) 09:40, 2 June 2007 (UTC)
- If AC fails, then there must be. Given a set X, if any two cardinalities are comparable, then there must be an injection from X into its Hartogs number (because there certainly can't be an injection in the reverse direction). Therefore X can be wellordered. Thus if all cardinalities are comparable, AC holds. --Trovatore 19:06, 2 June 2007 (UTC)
- Thanks for all that. I'm not sure I really understand all that, in particular the Hartog's number, I think I'll look at that more closely after my next exam (I hate it when real life interferes with Wikipedia :-) ).
- Returning to this article, I've reread your earlier comments and they are starting to make sense to me - I can see how AC is irrelevant to this article. And that there is no need for the article to take about "well-ordering" the cardinals. Aaron McDaid (talk - contribs) 20:50, 2 June 2007 (UTC)
- If AC fails, then there must be. Given a set X, if any two cardinalities are comparable, then there must be an injection from X into its Hartogs number (because there certainly can't be an injection in the reverse direction). Therefore X can be wellordered. Thus if all cardinalities are comparable, AC holds. --Trovatore 19:06, 2 June 2007 (UTC)
- I had thought that the Cantor–Bernstein–Schroeder theorem tells us that the cardinals can be ordered without using AC. But I'm not so sure now. It says that if |A|<=|B| and |B|<=|A|, then |A|=|B|. Was I mistaken in assuming that it's always true that |A|<=|B| and/or |B|<=|A| ? i.e. Without AC there may be a pair of cardinalities that cannot be ordered? Aaron McDaid (talk - contribs) 09:40, 2 June 2007 (UTC)
- I must have mixed up the two sections 'Statement and proof' and 'Discussion and consequences'. However, I still think their may be gaps in the article with regard to the axiom of choice (AC). Is it true that cardinals can ordered with or without AC? And that AC is only required to well-order them? I'm no longer implying there's a flaw in the proof, just being curious with a view to improving the article if appropriate. Aaron McDaid (talk - contribs) 08:55, 2 June 2007 (UTC)