= Transitive closure preserves set-theoretic rank
{title2=$\operatorname{rank}\operatorname{TC}(x)=\operatorname{rank}x$}
Using the convention that $\operatorname{TC}(x)$ is the least <transitive set> containing $x$ as a subset, put $x_0=x$, $x_{n+1}=\bigcup x_n$ and $\operatorname{TC}(x)=\bigcup_{n<\omega}x_n$. If $\operatorname{rank}x=\alpha$, every element encountered is an iterated member of an original element and has rank below $\alpha$. Thus the closure has rank at most $\alpha$, while $x\subseteq\operatorname{TC}(x)$ gives the reverse inequality. A convention requiring $x$ itself to be an element instead has rank $\alpha+1$.
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