How to Prove a Set Theory Ordinal Relationship?

[Fermat1]

Let $\beta$ be an ordinal.

Prove that $A\cap \bigcup\beta=\bigcup\{A\cap X\mid X \in \beta\}$

I'm not sure on this. It looks a bit like union distributing over intersection. Please help.

 

[Valkarie]

Let $A$ and $\beta$ be sets. We will show that $A \cap \bigcup \beta = \bigcup \{A \cap X \mid X \in \beta\}$. Let $x \in A \cap \bigcup \beta$. This means that $x \in A$ and $x \in \bigcup \beta$, the union of all sets in $\beta$. Since $x \in \bigcup \beta$, there must exist some $X \in \beta$ such that $x \in X$. Then, since $x \in A$ and $x \in X$, $x \in A \cap X$. Therefore, $x \in \bigcup \{A \cap X \mid X \in \beta\}$. Now let $x \in \bigcup \{A \cap X \mid X \in \beta\}$. This means that there exists some $X \in \beta$ such that $x \in A \cap X$. Since $x \in A \cap X$, it follows that $x \in A$ and $x \in X$. Since $x \in X$, $x \in \bigcup \beta$. Thus, $x \in A \cap \bigcup \beta$. We have shown that $A \cap \bigcup \beta \subseteq \bigcup \{A \cap X \mid X \in \beta\}$ and $\bigcup \{A \cap X \mid X \in \beta\} \subseteq A \cap \bigcup \beta$. Therefore, $A \cap \bigcup \beta = \bigcup \{A \cap X \mid X \in \beta\}$.