2. ## Re: Solve Sets

this is pretty straightforward if you just apply the definitions.

The general way to show two sets are equal is to show they are both subsets of one another, i.e. $A \subset B \wedge B \subset A \Rightarrow A=B$

I'll do one direction and leave the other to you.

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

then by definition $(x,y) \in (A \times B) \wedge (x,y) \in (C \times D)$

then again by definition this implies

$x \in A \wedge x \in C,~~y \in B \wedge y \in D$

thus $x \in A \cap C, ~~y \in B \cap D$ and thus $(x,y) \in (A \cap C) \times (B \cap D)$

thus $(A \times B) \cap (C \times D) \subset (A \cap C) \times (B \cap D)$

now you do the other direction.

3. ## Re: Solve Sets

Originally Posted by osamamansoor2016
Am I wrong, is the first question not, Prove that $\displaystyle A\times (B\setminus C)=(A\times B)\setminus (A\times C)\Large\color{red}??$
\displaystyle \begin{align*}(a,d)\in (A\times (B\setminus C)&\iff (a\in A) \wedge [d\in (B\setminus C)]\\&\iff (a\in A) \wedge[d\in B\wedge d\notin C]\\&\iff [(a\in A)\wedge (d\in B)] \wedge[d\in A\wedge d\notin C] \\&\iff (a,d)\in [A\times B]\setminus [A\times C]\end{align*}