Talk:Hartogs number

Invalid (or at least unclear) "Proof"

The proof given appears to be invalid or at least so unclear as to be unconvincing. JRSpriggs 06:14, 26 June 2007 (UTC)

I'm working on it. You could have tagged it with {{expert}}.... — Arthur Rubin | (talk) 18:53, 1 December 2007 (UTC)

Can we assume that, in ZF, if X is a set, then X × X is a set. The proof I'm familiar with in is NBG, where that's one of the finite set of axioms encompassing comprehension. — Arthur Rubin | (talk) 19:03, 1 December 2007 (UTC)

See Kripke–Platek set theory#Proof that Cartesian products exist for a proof that Cartesian products exist. Or use the axiom of powerset#Consequences and the axiom schema of specification. JRSpriggs 22:11, 1 December 2007 (UTC)

And thanks for reworking the proof, it is much clearer now. JRSpriggs 22:27, 1 December 2007 (UTC)

Greater than vs. not less than or equal

Forgive my ignorance, but this sentence confuses me:

If X cannot be wellordered, then we can no longer say that this α is the least wellordered cardinal greater than the cardinality of X, but it remains the least wellordered cardinal not less than or equal to the cardinality of X.

What is the difference between "greater than" and "not less than or equal to"? Solemnavalanche (talk) 19:52, 23 November 2008 (UTC)

See the article on cardinality. If A is a set of cardinality κ and B is a set of cardinality μ, then $\kappa \nleqslant \mu \,$ means that there is no injection from A to B while $\kappa >\mu \,$ means that there is an injection from B to A in addition to the absence of an injection from A to B. Does that clarify it? JRSpriggs (talk) 21:29, 23 November 2008 (UTC)

Thanks for the prompt response. It does clarify it somewhat, but it raises a further question. Does this mean that, absent the axiom of choice, there could exist pairs of sets for which no injection exists in either direction? That is, we could have $\kappa \nleqslant \mu \land \mu \nleqslant \kappa \,$? That clashes with my intuition, so I want to be sure I'm understanding you (and these articles) correctly. I see that the Law of Trichotomy article appears to say the same thing; but I'm having a difficult time imagining what such sets A and B would look like. Is there an article that addresses that question? Solemnavalanche (talk) 04:04, 24 November 2008 (UTC)

Well, it's good that it conflicts with your intuition, because the axiom of choice is intuitively correct.

But anyway, a good example is this: Let one of your sets be R, the set of all real numbers, and the other one be the Hartogs number of R. Then by definition there's no injection from the second to the first, and if there were an injection from the first to the second, that would imply that the reals could be wellordered.

Actually this generalizes, and shows that trichotomy implies the axiom of choice. --Trovatore (talk) 04:41, 24 November 2008 (UTC)