Verifying that $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$

Question

I believe that I have proved that $(A \times B) \cap (C \times D)= (A \cap C) \times (B \cap D)$, but I am not completely certain that every step I have written is reversible. Here is what I have: \begin{align*} x \in (A \times B) \cap (C \times D) & \iff x \in A \times B \text{ and } x \in C \times D \\ & \iff x = (a,b) \text{ for } a \in A, \; b \in B, \; \text{ and } x = (c,d) \text{ for } c \in C, \; d \in D\\ & \iff x = (a,b) = (c,d) \text{ for } a \in A, \; b \in B, \; c \in C, \; d \in D \\ & \iff x = (\alpha, \beta) \text{ where } \alpha \in A, \; \alpha \in C, \; \beta \in C, \; \beta \in D \\ & \iff x = (\alpha, \beta) \text { where } \alpha \in A \cap C, \; \beta \in C \times D \\ & \iff x \in (A \cap C) \times (C \times D) \end{align*} The first line is from the definition of intersetion. The second line is from the definition of the Cartesian product. The third line is from transitivity of equality. The fourth line is just a rewrite since the notion of an ordered tuple is well-defined. (I am not sure exactly how to say this other than that we clearly cannot have $x = (a,b)$ and $x = (c,d)$ where $a \neq c$, because then $x$ is a meaningless object.) The fifth line is from the definition of intersection. The sixth line is from the definition of the Cartesian product.

How is this?

Answer 1

At the very beginning of the proof, just write let $x = (\alpha, \beta)$ to avoid introducing $a, b, c,d$.

\begin{align} (\alpha, \beta) \in (A\times B) \cap (C \times D) & \iff (\alpha, \beta) \in (A\times B) \land (\alpha, \beta) \in (C \times D) \\ &\iff \alpha \in A \land \beta \in B \land \alpha \in C \land \beta \in D \\ &\iff \alpha \in (A \cap C) \land \beta \in (B \cap D)\\ & \iff (\alpha, \beta) \in (A \cap C) \times (B\cap D) \end{align}

Remark:

If $x=(a,b)$ and $x=(c,d)$, hence we have $(a,b)=(c,d)$ which means $a=c$ and $b=d$.

Answer 2

Close.

Notice that the expression equates an intersection of Cartesian products with a Cartesian product of intersections, so we should be dealing with sets of ordered pairs from the start, rather than trying to introduce the fact in the middle (and so avoid getting in a muddle).

$$\begin{align}\langle x,y\rangle\in ((A\times B)\cap (C\times D)) &\iff (\langle x,y\rangle\in (A\times B))~\land~(\langle x,y\rangle\in(C\times D))&&\text{Definition of Intersection}\\[1ex]&\iff \phantom{(x\in A\land y\in B)~land~(x\in C\land y\in D)}&&\text{Definition of Cartesian Product}\\[1ex]&\iff \phantom{(x\in A\land x\in C)~\land~(y\in B\land y\in D)}&&\text{Commutation and Association}\\[1ex]&\iff\phantom{ (x\in (A\cap C))\wedge (y\in (B\cap D))}&&\text{Definition of Intersection}\\[1ex]&\iff \langle x,y\rangle\in ((A\cap C)\times(B\cap D))&&\text{Definition of Cartesian Product}\\[4ex]\therefore\qquad(A\times B)\cap (C\times D)&= (A\cap C)\times(B\cap D) \end{align}$$