Prove that $A \times B \times (A \cap C) = (A \times B \times A) \cap (A \times B \times C)$

Question

I can't really do this question because I'm having trouble figuring out how I am supposed to work with the "$\times$" operation.

Answer 1

An element (i.e. a set) belongs to $A\times B\times (A\cap C)$ precisely when it is of the form $(a,b,x)$ for $a\in A, b\in B, x\in A, x\in C$ (as $x\in A\cap C$). An element belongs to $A\times B\times A$ and $A\times B\times C$ precisely when it is of the form $(a,b,x)$ for $a\in A, b\in B, x\in A, x\in C$, so the conditions are equivalent and the sets equal.

Answer 2

The $\times$ opperator is the Cartesian product, and in this case the triple product.

$$A\times B\times D = \{(x,y,z): x{\in}A, y{\in}B, z {\in}D\}$$

Thus an arbitrary element in $A\times B\times (A\cap C)$ is a triplet, let us say $(x,y,z)$, where the members are elements in $A, B,$ and $(A\cap C)$ respectively (by definition of Cartesian tripple product).   Then we ...say more stuff... that concludes that $A\times B\times (A\cap C) \subseteq (A\times B\times A)\cap(A\times B\times C)$.

Likewise an arbitrary element in $(A\times B\times A)\cap(A\times B\times C)$ is an element both in $(A\times B\times A)$ and in $(A\times B\times C)$ (by definition of set union).   We then ...say even more stuff... and conclude the converse; that $A\times B\times (A\cap C) \supseteq (A\times B\times A)\cap(A\times B\times C)$.

And we would be done.