In the book<>written by Charles C.Pinter it said that in order to prove "the axiom of replacement together with strengthened version of axiom of choise imply axiom of limitation of size"(X is a proper class iff it is 1-1 correspondence with the universal sets V

In my perspective,I can't understand why it need the 'strong' version of axiom of choice.I think axiom of replacement with usual axiom of choice can imply VN sufficiently.Here is my proof:

1.It is obvious that X→V is injective(By definition V contains all the sets).Thus INJ holds

2.Now we shall show X→V is surjective:

For any Va∈V('a'is any ordinal number and Va is a set),by axiom of replacement it follows that there exists a set A⊂X which is A→Va is bijective.If such a set 'A'does not exist then X would be a set.A contradiction.Thus SUJ holds.

Therefore X→V is 1-1 correspondence.

I haven't use strong version of AC.Can someone tell me is there any mistake or logic lackage in my proof(especially for proving X to V is surjective) and give me an explicit correct proof for this exercise?Thanks!


closed as unclear what you're asking by Andrés E. Caicedo, JonMark Perry, ml0105, hardmath, Leucippus Jul 20 '16 at 0:12

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    $\begingroup$ You might want to improve your question in the following ways: (1) Make the question body self-contained, and not relying on the title for the actual question. (2) Include the formulation of the axiom of global choice and the name of the book. (3) Use $\rm\LaTeX$ (or rather MathJax) to format your question if you are familiar with that typesetting system. $\endgroup$ – Asaf Karagila Jul 19 '16 at 9:05
  • $\begingroup$ Sorry.The book's name is called <<A Book of Set Theory>>.I don't know why i forgot. $\endgroup$ – majin vegeta Jul 19 '16 at 13:18
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    $\begingroup$ You should use the edit feature to edit this into your question. $\endgroup$ – Asaf Karagila Jul 19 '16 at 15:12
  • $\begingroup$ en.wikipedia.org/wiki/Haddocks%27_Eyes $\endgroup$ – Andrés E. Caicedo Jul 19 '16 at 15:26
  • $\begingroup$ I followed you down to "If such a set 'A' does not exist then X would be a set." Nothing you said to that point strikes me as supporting that conclusion, so that might be where you edit after the improvements @Asaf already suggested. I suspect the issue here is that the usual (weak?) axiom of choice only allows a choice function for set-collections of nonempty sets, so creating a bijection $X \to V$ as proposed would be out of bounds. $\endgroup$ – hardmath Jul 19 '16 at 17:18

For every $\alpha$ there is a proper class of subsets of $X$ which can be mapped onto $V_\alpha$, and for each of those there are many surjections. You need to choose a set for each $\alpha$ simultaneously and choose a surjection, and do it in a coherent way.

Not to mention, that in order to prove that there is a bijection between $X$ and $V$ you need to produce a bijection, or two injections. In order to prove that a surjection from $X$ onto $V$ can be reversed to provide you with an injection from $V$ into $X$ you need to literally choose a preimage for each set. That is exactly where the global choice axiom comes in.

In order to prove the principle, suppose that $X$ is a proper class, then by the axiom of replacement if $f\colon A\to X$ is a function with $A$ being a set, then $\operatorname{rng}(f)$ is a set, and therefore not $X$ itself. Now by the axiom of choice, every two sets are comparable. So in particular every set must have an injection into some sufficiently large $X\cap V_\alpha$.

We construct an injection from $V$ into $X$ by induction, fix a choice function from the class of all non-empty sets. Suppose we constructed an injection $f_\alpha$ from $V_\alpha$ into $X$. Let $\beta$ large enough such that there is an injection from $V_{\alpha+1}\setminus V_\alpha$ into $V_\beta\cap X\setminus\operatorname{rng(f_\alpha)}$. We choose an injection $g_\alpha$ like this from the set of possible injections. Now define $f_{\alpha+1}=f_\alpha\cup g_\alpha$.

For limit cases, and $\alpha=\rm Ord$ take $f_\alpha=\bigcup_{\beta<\alpha}f_\beta$.

I will leave it to you to see that $f_\alpha$ is always an injection and for $\alpha=\rm Ord$ it is an injection from with domain $V$ into $X$.

  • $\begingroup$ Therefore what you mean is my proof is correct.The only thing is the proof of the statement"For every α there is a proper class of subsets of XX which can be mapped onto Vα, and for each of those there are many surjections"is a consequence of axiom of global choice which I didn't mention.Is it?Also can you tell me is there any logic mistake I made?Or the only lackage is what you have noticed? $\endgroup$ – majin vegeta Jul 19 '16 at 13:23
  • $\begingroup$ I mean that your proof is correct if you appeal to global choice. If you want not to appeal to global choice, how can you choose for each $\alpha$ an appropriate $X_\alpha$ and $f_\alpha$, and ensure that the union $\bigcup f_\alpha$ is even a function (you need to somehow ensure that you get a function)? $\endgroup$ – Asaf Karagila Jul 19 '16 at 15:11
  • $\begingroup$ Therefore if I use the global choice axiom at where you mentioned then my proof is perfect?Sorry I am really concentrating on my process. $\endgroup$ – majin vegeta Jul 19 '16 at 15:36
  • $\begingroup$ Also I have another question:why lots of related set theory books do not state the axiom of global choice?Is it still not be wide-accepted by mathematicians? $\endgroup$ – majin vegeta Jul 19 '16 at 15:38
  • $\begingroup$ You still need to argue how you can ensure you choose these functions to be coherent. For your second question, because mathematicians are usually interested in sets, so class choice is excessive and unnecessary. It also requires us to be able and talk about classes, which means going beyond ZFC for better and for worse. $\endgroup$ – Asaf Karagila Jul 19 '16 at 15:39

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