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I'm trying to prove the following lemma left as an exercise in my course notes.

If $x$ is a well-founded set, then $$ \mathrm{rank}(x) = \sup \{ \mathrm{rank}(y) + 1 : y \in x \} $$

Proof. Let $\alpha = \sup \{ \mathrm{rank}(y) + 1 : y \in x \}$. It suffices to show that $x \in V_{\alpha + 1}$ but $x \notin V_\alpha$. Notice that for any $y \in x$, $\mathrm{rank}(y) < \alpha$ by definition. Hence, $y \in V_\alpha$. Since $\forall y \in x : y \in V_\alpha$, $x \subseteq V_\alpha$, so by definition $x \in V_{\alpha + 1}$.

The part I'm struggling to show is that $x \notin V_\alpha$. I don't see how assuming this leads to a contradiction.

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    Some texts (e.g. Kunen : Set Theory:An Introduction to Independence Proofs) define $rank(x)$ to be $\sup \{rank (y)+1:y\in x\}.$ What def'n do you use? – 2017-04-13

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As $\alpha$ is the least ordinal that is $>\operatorname{rank}(y)$ for all $y\in x$, the assumption that $\operatorname{rank}(x)<\alpha$ means that $\operatorname{rank}(x)\le\operatorname{rank}(y)$ for some $y\in x$. But that is impossible.

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    How is it that assuming $x \in V_\alpha$ implies $x < \alpha$? I have a lemma that shows that $x \in y \implies \operatorname{rank}(x) \leq \operatorname{rank}(y)$, but that isn't strong enough by itself. – 2017-02-06
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Suppose $y \in BF(\alpha)$ then by definition $rank(y) < \alpha$.

Now return to y. You know that $x \in y \rightarrow rank(x) < rank(y)$. Then $$ rank(x) + 1 \leq rank(y) \mathbb \space \forall x\in y $$

Now look back at your definition of supremum. With the above line we have created an upper bound lower than the supremum $a$ which contradicts the minimality of $\alpha$. Thus $ y \not\in BF(\alpha)$