If $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$

Question

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 $a_x = x$.

Answer 1

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 $y\in a_x$.

Answer 2

In my original post, I wrote $a_x=\{y\in a:y\subset x\}$, but the definition Cameron actually gives is $a_x=\{y\in a:y < x\}$. Does this make a difference?

Either way, it seems that:

$$\begin{align} a_x &= \big\{y\in a:y<x\big\} \\ &= \big\{y\in x\cup\{x\}:y<x\big\} \\ &= \big\{y\in x:y<x\big\}\cup\big\{x\in\{x\}:y<x\big\} \\ &= x\,\cup\,\varnothing \\ &= x \end{align}$$

where $\big\{x\in\{x\}:y<x\big\} = \varnothing$ because $x\nless x.$