If $X$ is a set of ordinals then so is $\bigcup X$

Question

Theorem. If $X$ is a set of ordinals then so is $\bigcup X.$

Claim 1. $\bigcup X$ is well-ordered.

Proof of Claim 1. Let $Y\subseteq\bigcup_{I\in X}I.$ There is a $I\in X$ such that $Y\subseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $\bigcup_{I\in X}X$ is well-ordered.

Claim 2. If $x\in\bigcup X$, then $x\subseteq\bigcup X.$

Proof of Claim 2. Let $x\in\bigcup X.$ So there is a $I\in X$ such that $x\in X.$ Since $X$ is an ordinal, $x\subseteq X\subseteq \bigcup X$, so $x\subseteq\bigcup X.$

Proof of theorem. By the Claim 1 and the Claim 2.

May you check my proof? Thanks...

Answer 1

So, for claim 1, let $Y\subseteq \bigcup X$, then it is not true in general that $Y\subseteq I$ for some $I\in X$.
However, it's true for any element $y$ of $Y$: $y\in I$ with some $I\in X$.
Now we can take the smallest element $y_0$ in $Y\cap I\, \subseteq I$, and prove it's the smallest element of $Y$.

The proof of claim 2 is correct.