Show that $(A\cup B)\times C=(A\times C)\cup (B\times C)$

Question

This is my try attempt:

22) ${\text{Show that }(A\cup B)\times C=(A\times C)\cup (B\times C)]\\ \text{We take some }{(x,y)\in[(A\times C)\cup(B\times C)]\\ = [x\in A\wedge y\in C]\vee[x\in B\wedge y\in C]\\ = [x\in A\vee x\in B]\wedge[y\in C]\\ = [x\in (A\cup B)]\wedge[y\in C]\\ = (x,y)\in (A\cup B)\times C }}$

Please tell me if I am wrong.

Answer 1

You've shown that $(A\times C)\cup(B\times C)\subseteq (A\cup B)\times C$. You need to prove the other direction as well, for equality.