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!
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 $X$.