$(\bigcup_{\alpha \in J} (U_\alpha \times V_\alpha)) \cap (A \times B) = \bigcup_{\alpha \in J}((U_\alpha) \cap A) \times (V_\alpha \cap B))$

Question

I'm trying to prove that if $X$ and $Y$ are sets and $A \subset X$ and $B \subset Y$. Then for every element $\alpha \in J$ in an indexing set $J$. If you let $U_\alpha$ be a subset of $X$ and $V_\alpha$ be a subset of $Y$ then: $(\bigcup_{\alpha \in J} (U_\alpha \times V_\alpha)) \cap (A \times B) = \bigcup_{\alpha \in J}((U_\alpha) \cap A) \times (V_\alpha \cap B))$. I'm not really sure where to start and any help would be appreciated.

Answer 1

If $(x,y)$ is a point in $\left(\bigcup_{\alpha \in J} (U_\alpha \times V_\alpha)\right) \cap (A \times B)$, this means that $(x,y) \in A \times B$, so $x \in A$ and $y \in B$ and also that there is ome $\alpha \in J$ so that $(x,y) \in U_\alpha \times V_\alpha$, which means that for that same $\alpha$ we have $x \in U_\alpha$ and $y \in V)\alpha$; it follows then that $x \in U_\alpha \cap A$ and and $y \in V_\alpha \cap B$ so that $(x,y) \in (U_\alpha \cap A) \times (V_\alpha \cap B)$ and so $(x,y) \in \bigcup_{\alpha \in J} (U_\alpha \cap A) \times (V_\alpha \cap B)$, as required.

All the steps are in fact reversible, so taking a point $(x,y) \in \bigcup_{\alpha \in J} (U_\alpha \cap A) \times (V_\alpha \cap B)$ we can quite easily sow it to be in $\left(\bigcup_{\alpha \in J} (U_\alpha \times V_\alpha)\right) \cap (A \times B)$ as well, and we've then shown both inclusions, hence set equality.