Why is it not obvious that the Zermelo hierarchy is a hierarchy?

57 Views Asked by At

In Sets, Logic and Categories, Cameron writes

First, [we show that the Zermelo hierarchy] really is a hierarchy: the sets get larger as we progress. (This is not obvious; each set is the power set of its predecessor, and most sets $X$ don't satisfy $X\subseteq\mathcal{P}X$.)

I'm stuck on the parenthetical sentence. What does he mean by "each set is the power set of its predecessor"? (Is this referring to ordinals, or to sets in general?) And what does he mean by "most sets $X$ don't satisfy $X\subseteq\mathcal{P}X$"? (Most sets = most ordinals? Certainly this holds for finite sets, no? But not ordinals?)

1

There are 1 best solutions below

1
On BEST ANSWER

For "each set is the power set of its predecessor", it refers to the hierarchy itself. If $V_\alpha$ is the stage $\alpha$ of the hierarchy, then $V_{\alpha+1}=\mathcal{P}(V_\alpha)$.

For "most sets $X$ don't satisfy $X⊆\mathcal{P}(X)$", you can check that indeed finite sets do not satisfy this except for very special cases. For instance, if $X$ is a singleton, then $\mathcal{P}(X)=\{X,\emptyset\}$, and unless $X=\{X\}$ (impossible with the foundation axiom) or $X=\{\emptyset\}$, you don't get $X\subset \mathcal{P}(X)$. (The good news is that $V_1$ is indeed $\{\emptyset\}$, so there is no contradiction.)