If $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$ I wish to show that $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$ (with $y<x$ for all $y\in x$).
I have seen a proof in Cameron's Sets, Logic and Categories (p.41), but I find it unsatisfactory. He writes:

The set $a = x\,\bigcup\,\{x\}$ (with order as specified in the the statement of the theorem) has as sections all the sections of $x$ and one additional one, namely $a_x$. But since all the elements of $x$ are smaller than $x$, we have $a_x = x$.

I don't see why it necessarily follows from this that $a_x = x$. By definition, $a_x = \{y\in a : y\subset x\}$, so $a_x\subseteq x$, but this is different than $a_x = x$.

Suppose $y\in x$.  Since $x$ is an ordinal, this implies $y\subset x$.  So, $y$ is an element of $a$ such that $y\subset x$, which by definition means that $y\in a_x$.
