Showing that $\bigcap A$ is the least element for the set $A$ where $A$ is a set of ordinals.

Question

The notes I am reading define a set $x$ to be an ordinal provided $x$ is transitive and every element in $x$ is transitive.

Let $A$ be a set of ordinals. I have shown that $\bigcap A$ is an ordinal. I having trouble showing that $\bigcap A \in A$.

I realized this is obvious in the finite case. Let $x$ and $y$ be ordinals. Then $\bigcap \{x, y\} = x \cap y$. Since ordinals are linear ordered(which I have shown), then $x \in y$ or $x = y$ or $y \in x$. Suppose $x \in y$, then I can show that $x \cap y = x$.

Sorry if this a dumb question since this seems simplee.

Answer 1

Perhaps one way to do it is to note that $\alpha$ and $\beta$ are ordinals, and $\alpha\subseteq \beta$, then either $\alpha=\beta$ or $\alpha\in\beta$. For example, see here.

You know that $\cap A$ is an ordinal. We know that $\cap A\subseteq \beta$ for all $\beta\in A$. If $\cap A$ is equal to some $\beta$ in $A$, then we are done. Otherwise, we have $\cap A\subsetneq \beta$ for all $\beta\in A$, and therefore, by the result noted above, $\cap A\in\beta$ for all $\beta\in A$.

But that would mean that $\cap A\in\beta$ for all $\beta\in A$, and therefore that $\cap A\in\cap A$... which is impossible for ordinals.