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

  • $\begingroup$ I think that you have another account, no? You can ask on the meta site to have them merged. $\endgroup$ – Asaf Karagila Feb 23 '13 at 14:39
  • $\begingroup$ 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 $V_\alpha$? $\endgroup$ – Adam Epstein Feb 23 '13 at 15:44
  • $\begingroup$ Upvote for a compelling statement that requires Replacement. $\endgroup$ – Adam Epstein Feb 23 '13 at 15:48
  • $\begingroup$ 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 $V_{\alpha_i}\cap A\setminus A\cap\bigcup_{j<i}V_{\alpha_j}$. $\endgroup$ – Asaf Karagila Feb 23 '13 at 15:52
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 23 '13 at 15:53

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.

  • $\begingroup$ 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!! $\endgroup$ – Asaf Karagila Feb 23 '13 at 15:47
  • $\begingroup$ Thanks, Asaf. But these are in Jech's The Axiom of Choice rather than Set Theory $\endgroup$ – Adam Epstein Feb 23 '13 at 15:54
  • $\begingroup$ Adam, that's what I edited. Yes. $\endgroup$ – Asaf Karagila Feb 23 '13 at 16:16
  • $\begingroup$ 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 $\endgroup$ – Gérard Lang Feb 23 '13 at 17:58
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 23 '13 at 18:03

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