5
$\begingroup$

We work in the set theory NBG with the axiom of (local choice but without global (class) choice. For every class A P(A) is the class of all sets x included in the class A.

We know that P(A) is a set iff A is a set and a proper class iff P(A) is a proper class. We also know that if A is a set there is no bijection between A and P(A), and that P(P(V))=V, where V is the universal class. It is clear that if there is a bijection between A and V, then there is a bijection between A and P(A).

Question: Is it true that if there is a bijection between A and P(A) then there is a bijection between A and V?

This question is not interesting under global choice, where all proper classes are bijective.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

Yes, this is provable in NBG. To see this, let $F$ be a one-one function from $\mathcal P(A)$ into $A$. By transfinite recursion on $\in$, we define a function $G$ from $V$ to $A$ such that $G(x) = F(G[x])$. A simple induction then establishes that $G$ is one-one.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Dear Sam, let me thank you very much, first to have remarked that Asaf's answer was not sufficient and moreover to give a definite answer to my question. As I understand it, suppose G(α) is a function from V(α) into A and let X∈ V(α+1)/V(α). $\endgroup$
    – Gérard Lang
    Commented Dec 6, 2014 at 15:51
  • $\begingroup$ oops. I continue. We have that y∈x implies y∈V(α), so that G[x]={G(y)/y∈x} is a well-defined set with G[x]⊆V(α)⊆P(A). $\endgroup$
    – Gérard Lang
    Commented Dec 6, 2014 at 15:58
  • $\begingroup$ @GérardLang That's right; we can also recursively define an ordinal sequence of functions -- so that $G(\alpha)$ is a one-one function from $V_\alpha$ into $A$, $G(\alpha+1)$ extends $G(\alpha)$ to $V_{\alpha+1}$ uniquely in the way I suggest, and $G(\lambda) = \bigcup_{\alpha<\lambda} G(\alpha)$ for $\lambda$ a limit. A simple induction then shows that each $G(\alpha)$ is one-one, and thus that $\bigcup_{\alpha\in On} G(\alpha)$ is a one-one function from $V$ into $A$ as required. $\endgroup$
    – Sam Roberts
    Commented Dec 6, 2014 at 19:22
3
$\begingroup$

It suffices to show that for $A=\sf Ord$. Since in that case global choice holds and the rest follows.

Fix a pairing function for ordinals, namely a bijection between $\sf Ord$ and $\sf Ord\times Ord$. Now given a bijection between $\sf Ord$ and $\cal P\sf (Ord)$, pick for each $\alpha$ the least $X_\alpha\in\cal P\sf (Ord)$ such that $X_\alpha$ encodes the structure $(V_\alpha,\in)$, and $X_\alpha$ is disjoint from $\bigcup_{\beta<\alpha} X_\beta$. Namely a subset of the ordinals which, when translated using the pairing function, gives us a structure which is extensional and well-founded, and its transitive collapse is $V_\alpha$.

Now we have a bijection between the ordinals and the universe, defined by the following, given $x$ of rank $\alpha$, $x$ is mapped to the unique $\eta\in X_\alpha$ which encodes $x$.

(Alternatively, you don't need to require that the $X_\alpha$'s are disjoint, and this defines a surjection from $\sf Ord$ onto $V$ in a similar way.)

Improve this answer
$\endgroup$
6
  • $\begingroup$ Dear Asaf, Thank you very much for this nice answer. If I correctly understand, the negation of the axiom of global choice is equivalent with the existence of a proper class A such that A is bijective with P(A) and not bijective with V. And by the model built by Ali Enayat, where On, P(On) and P(P(On)) are not bijective, it is possible that such A is neither bijective with On, nor with P(On). Would it be possible that the class W defined by JD Hamkins be such a class ? $\endgroup$
    – Gérard Lang
    Commented Dec 6, 2014 at 11:08
  • $\begingroup$ You just need to notice that $\mathcal P\sf (Ord)$ can always be linearly ordered. In a model where there is a class which cannot be linearly ordered, this class cannot be put into bijection with $\mathcal P\sf (Ord)$; and if this class is such that there is no injection from $\sf Ord$ into it, then this class cannot be put into bijection with $V$ either. Joel Hamkins suggested such model in an answer to a question of mine, to which I linked in one of your recent questions. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 6, 2014 at 11:28
  • 2
    $\begingroup$ @AsafKaragila Sorry to be dense, but how does the general claim (for all $A$, if $A$ is bijective with $\mathcal P(A)$, then $A$ is bijective with $V$) follow from the instance $A = On$ (if $On$ is bijective with $\mathcal P(On)$, then $On$ is bijective with $V$)? $\endgroup$
    – Sam Roberts
    Commented Dec 6, 2014 at 13:49
  • 1
    $\begingroup$ @Gerard: As Sam points out, I misread the question. Please unaccept my answer. I will leave it for a while, if only to keep the comments below it visible. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 6, 2014 at 15:08
  • 3
    $\begingroup$ @Gerard: I should have been more careful. I read the question as asking about "Every class $W$ has a bijection with $\mathcal P(W)$", which is what my answer shows. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 6, 2014 at 15:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .