Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a proof in ZFC, using AC and the axiom of foundation, that given any proper class A, every set can be injected into A. I wonder if we could have a proof of this that does not use Foundation. Gérard Lang

share|improve this question
    
I think that you have another account, no? You can ask on the meta site to have them merged. –  Asaf Karagila Feb 23 '13 at 14:39
    
Interesting. Would it be correct to infer that the overall strategy is to prove, presumably by transfinite induction, that it is possible to inject any $V_\alpha$? –  Adam Epstein Feb 23 '13 at 15:44
    
Upvote for a compelling statement that requires Replacement. –  Adam Epstein Feb 23 '13 at 15:48
    
Adam, in ZFC (proper!) it's easy to prove. Let $x$ be a set, well order it. Since $A$ is a proper class for every $\alpha$ there is $\beta>\alpha$ such that $V_\beta\cap A$ has more elements than $V_\alpha\cap A$. Let $\alpha_i$, $i<|x|$ be a strictly increasing sequence of the least ordinals where information on $A$ is added. For every $i<|x|$ map $x_i\in x$ to some element in $V_{\alpha_i}\cap A\setminus A\cap\bigcup_{j<i}V_{\alpha_j}$. –  Asaf Karagila Feb 23 '13 at 15:52
    
Also, it should be remarked that the use of choice is also essential. It is consistent that there is a proper class without a countably infinite subset. –  Asaf Karagila Feb 23 '13 at 15:53

1 Answer 1

up vote 5 down vote accepted

No. This principle is known as The Injection Principle

See in Jech Axiom of Choice, Chapter 9, Problems 3 and 4 both give us a models of ZF+Atoms (so foundation fails) with choice in which the injection principle fails.

share|improve this answer
    
Adam, neither of the editions. I don't know why I wrote "Set Theory". I was literally holding the book in my hand when I typed that answer. Strange... :-) Thanks!! –  Asaf Karagila Feb 23 '13 at 15:47
    
Thanks, Asaf. But these are in Jech's The Axiom of Choice rather than Set Theory –  Adam Epstein Feb 23 '13 at 15:54
    
Adam, that's what I edited. Yes. –  Asaf Karagila Feb 23 '13 at 16:16
    
Dear Asaf, thank you very much, I am happy to credit you with for this answer. But, in fact, I was thinking to ZFC set theory without atoms; does the result in Jech transfer without atoms ? It is perfectly true that I have two distinct counts in Mathoverflow for reasons that I cannot understand. My older one, with more than 500 points recognized my true e-mail beginning with"gerard_lang@". But, now it is impossible for me to connect on this count, and I have another newer one, with 1OO points, that wants only to recognize "gerard-lang@ ", that is not my correct electronic adress ! Gérard Lang –  Gérard Lang Feb 23 '13 at 17:58
    
Dear Gerard, If you have a proper class of sets of the form $\lbrace x\rbrace = x$ then I believe that you can treat them as atoms and repeat all the constructions and have this. –  Asaf Karagila Feb 23 '13 at 18:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.