Timeline for When do we have a bijection between a proper class A and its power set class P(A)?

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Dec 7, 2014 at 15:20	history	edited	Sam Roberts	CC BY-SA 3.0	deleted 39 characters in body
Dec 6, 2014 at 19:22	comment	added	Sam Roberts		@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.
Dec 6, 2014 at 15:58	vote	accept	Gérard Lang
Dec 6, 2014 at 15:58	comment	added	Gérard Lang		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).
Dec 6, 2014 at 15:51	comment	added	Gérard Lang		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).
Dec 6, 2014 at 14:42	history	answered	Sam Roberts	CC BY-SA 3.0