Question about the Von Neumann Rank of elements of sets Let $X$ be a transitive set in the Von Neumann Heirchary and let $rank(x)=\alpha$. Prove that if $\beta<\alpha$, then there exists $Z\in X$ with $rank(Z)=\beta$. 
My attempt is to consider $Z=\{z\in X | rank(z)<\beta\}$. Clearly $rank(Z)=\beta$, since for all $z\in Z$, $rank(Z)<\beta$, so $Z\subseteq V_\beta$. I'm having trouble showing that in fact $Z\in X$. 
Is this even the correct choice of $Z$? What does $X$ being transitive have to do with anything?
 A: The set $Z$ you've defined won't work... it is not necessarily true that $Z\in X$ (so it is good that you failed to show it).
Let $X$ have minimal rank $\alpha$ such that the property fails, and let $\beta < \alpha$ such that there is no set in $X$ with rank $\beta.$ If every set in $X$ had rank less than $\beta,$ then $X$'s rank wouldn't be greater than $\beta,$ and we've assumed there aren't any sets in $X$ with rank equal to $\beta,$ so there must be a set $y\in X$ with rank greater than $\beta.$ Let $t=\operatorname{trcl}(y).$ Then $\operatorname{rank}(t)=\operatorname{rank}(y) < \alpha$ and $t$ is transitive, so by minimality of $\alpha,$ $t$ contains a set of rank $\beta.$ But since $X$ transitive, $t\subseteq X,$ so $X$ contains a set of rank $\beta$ after all.
The intuition here is that a transitive set is 'packed down' as far as possible, so there can't be any missing ranks. This is made precise by the Mostowski collapse, which maps any set onto a unique transitive set in a way that respects the membership relation. The Mostowski collapse of a set $x$ recursively moves the sets in $x$ to the lowest possible rank consistent with preserving the membership relation on $x$, starting from the bottom (i.e. the $\in$-minimal elements of $x$, which it maps to the empty set). 
It can be expressed recursively as $\pi(y)=\{\pi(z): z\in y\cap x\},$ for $y\in x.$ And if $x$ is transitive, $x\cap y = y$ and so by induction, $\pi(y)=y$ and the collapse function is the identity.
A: Suppose that $\beta<\alpha$ and no element of $X$ has rank $\beta$. Since 
$$\operatorname{rank}(X)=\sup\{\operatorname{rank}(x)+1:x\in X\}\;,$$
we must have $\alpha>\beta+1$ and $Y=\{x\in X:\operatorname{rank}(x)>\beta\}\ne\varnothing$. Fix $y\in Y$ of minimal rank, and let $\gamma=\operatorname{rank}(y)$. Then $\beta<\gamma=\sup\{\operatorname{rank}(x)+1:x\in y\}$, and there is no $x\in y$ of rank $\beta$, since $X$ is transitive. As for $X$ above it follows that there is a $z\in y$ such that $\operatorname{rank}(z)>\beta$. But then $\beta<\operatorname{rank}(z)<\gamma$, contradicting the choice of $y$.
