$(C \times D) \cap (A \times B) \neq \emptyset \implies C \cap A \neq \emptyset$ and $D \cap B \neq \emptyset$

Question

If $C$ and $D$ are open sets, and $(C \times D) \cap (A \times B) \neq \emptyset$, then why is it necessarily true that $C \cap A \neq \emptyset$ and $D \cap B \neq \emptyset$?

Answer 1

Let $(x,y) \in (C \times D) \cap (A \times B)$.

Then $(x,y) \in C \times D$ and $(x,y) \in A \times B$.

That is $x \in C$ and $y \in D$ and $x \in A$ and $y \in B$.

Hence, $x \in C \cap A$ and $y \in D \cap B$.

Therefore $C \cap A \not = \emptyset$ and $D \cap B \not = \emptyset$.

Observe that this is true for any sets $A,B,C$ and $D$ as Milo Brandt said.