I'm working through Devlin's "The Joy of Sets". Theorem 1.7.7 states: Let $X$ be an ordinal. Let $Y\subset X$. If $Y$ is an ordinal, then $Y=X_{a}=\{x\in X\vert x \lt a\}$ for some $a \in X$.
Now, I know this theorem to be true but the first line of his proof seems wrong to me. I'm sure the misunderstanding is mine, but might you help me understand where I'm confused? He begins by letting $a$ be the smallest member of $X-Y$ and then simply states "Thus $X_{a}\subseteq Y$". I don't see how this follows. It seems to me that if $a$ is the smallest member of $X-Y$ then $Y \subseteq X_{a}$. I know that these two aren't mutually exclusive, but I don't get Devlin's intuition that $X_{a} \subseteq Y$.
To show you where I'm coming from, it seems to me that if $a$ is the smallest member of $X-Y$ then $a \notin Y$. However, $a\in X$ and so we know that $X_{a}$ is an ordinal and $X_{a} \subset X$. Thus both $Y$ and $X_{a}$ are ordinals that are subsets of $X$. Since $X_{a}=a$ and $a\notin Y$ we know that $X_{a}\notin Y$. This can only happen if $X_{a} = Y$ or if $Y \subset X_{a}$, thus we conclude $Y \subseteq X_{a}$.
Can someone explain to me where I've gone wrong so that I can understand why Devlin states "Thus $X_{a}\subseteq Y$". Thanks!
If $b<a$, then $b\not\in X-Y$ since $a$ is defined to be the smallest element of $X-Y$.
But if $b\not\in X-Y$ then necessarily $b\in Y$, and since $b$ was an arbitrary element of $X_a$ we obtain $X_a\subseteq Y$.