$\bigcup X$ = $\sup(X)$ for a set $X$ of ordinals

Question

Let $X$ be a set of ordinals. Prove that $\bigcup X$ = $\sup(X)$(which is the least ordinal $\ge$ every ordinal in $X$).

Answer 1

First we will check that $\bigcup X$ is also ordinal. To check this fact, we will prove that $\bigcup X$ is transitive and well-ordered with respect to $\in$.

If $y\in \bigcup X$ for all $y\in Y$, then there is some $x\in X$ satisfy that $y\in x$. By transitivity of $x$, we get $\bigcup y\in x$ so $\bigcup y \in\bigcup X$ (because $x\subset \bigcup X$.) Therefore $\bigcup X$ is transitive.

Let $y,z\in\bigcup X$, then there is $x\in X$ such that $y,z\in x$ (because for every ordinal $\alpha$ and $\beta$ satisfy $\alpha\subset\beta$ or $\beta\subset \alpha$.) Since $y$ and $z$ are elements of ordinal so $x\in y$ or $x=y$ or $y\in x$ holds.

Finally, we will check that $\bigcup X$ is well-ordered by respect to $\in$. Let $A$ be a nonempty subset of $\bigcup X$. Since $A$ is nonempty, there is = $x\in X$ that intersects to $x$ (that is, $A\cap x\neq\varnothing$.) Consider $\alpha=\min (A\cap x)$ (this minimum exists because $A\cap x\subset x$ and $x$ is ordinal.)

Let $\beta \in A$. If $\beta\in x$ then definition of $\alpha$ we get $\alpha=\beta$ or $\alpha=\beta$. If not, we get $x\in\beta$ or $x=\beta$ (because $x$ and $\beta$ are ordinal.) In that cases we get $x\subset \beta$, so $\alpha\in\beta$. So $\alpha$ is the minimal element of $A$.

Therefore $\bigcup X$ is ordinal and $x \subset \bigcup X$ for all $x\in X$. If $x\subset y$ for all $x\in X$, then by transitivity of $y$ we get $\bigcup X\subset y$. Therefore $\bigcup X=\sup X$.

Answer 2

Suppose first that $X$ has a maximal element $\mu$; then $\alpha\subseteq\mu$ for all $\alpha\in X$, so $\bigcup X=\mu$, and clearly $\mu=\sup X$.

Now assume that $X$ has no maximal element, and let $\mu=\sup X$. If $\xi\in\bigcup X$, there must be some $\alpha\in X$ such that $\xi\in\alpha$; and $\alpha\le\mu$, so $\alpha\subseteq\mu$, and therefore $\xi\in\mu$. Thus, $\bigcup X\subseteq\mu$. Now suppose that $\xi\in\mu$. Then $\xi<\mu$, so there is an $\alpha\in X$ such that $\xi<\alpha$, and hence $\xi\in\alpha\subseteq\bigcup X$. Thus, $\mu\subseteq\bigcup X$, and we have $\bigcup X=\mu$ in this case as well.