Union of a set of ordinals

Question

Let $x$ be a set of ordinals, and let $U_x:=\bigcup x$ be the union of all members of $x$. How do I prove $U_x$ is well-ordered under $\in$? I know how to do it if I can find an ordinal $\alpha$ such that $x\subseteq\alpha$, because then any $S\subseteq x$ will be a subset of $\alpha$ and hence, if it is nonempty, have a minimal element under $\in$. However, I cannot, in general, prove such an $\alpha$ exists. I can do it if $x$ is finite. How can I do it for infinite $x$s?

Answer 1

The trick is simply to limit the ordinals of the subset $S$ from above. Here is how we can prove this.

Take $S\subseteq\bigcup x$. There must be at least one $y\in x$ such that $y\cap S\neq\varnothing$, given $S\subseteq\bigcup x$, unless of course $S=\varnothing$ which we assume against. Pick one such $y$. There must be a minimal element $z\in y\cap S$, since $y$ is well-ordered and $y\cap S\neq\varnothing$. For all $t\in y\cap S:t\neq z$, we have $z\in t$. $z\in y$ because $z\in y\cap S\subseteq y$. If $y\in t\in S$, by transitivity of $t$ we will have $z\in t$. Thus, $z$ is the minimal element of $S$, which proves $\bigcup x$ is $\in$-well-ordered.