My attempt at proving $A\cup B=A \cup (B \cap A^{c})$

Question

This is my first time doing a proof where you have to show containment both ways. Is this proof correct?

Answer 1

Firstly, not that you use it, but it's a bad idea for your first assumption to be the very thing you're trying to prove!

Secondly, you cannot say that, if $x \in B$, then $x \in B \cap A^c$. You've just assumed that $x \in B$, but it might also be true that $x \in A$ as well. What you should do is assume $x \notin A$. Then, since $x \in A$ or $x \in B$, then we must have $x \in B$. In this case, we have $x \in B \cap A^c$.

Otherwise, the proof looks fine!

Answer 2

Start with the more complicated side (the right hand side in this case) and use the distributive and complement laws to get to the simpler side (the left hand side in this case) as follows:

$$A \cup (B \cap A^c)=(A \cup B) \cap (A \cup A^c)=(A \cup B) \cap U=A \cup B$$, where $U$ is the universal set (so $A,B \subseteq U$).

Answer 3

Your first step is wrong; $x\in(A\,\cup\,B)$ means that $x$ is in either or BOTH, hence when you say if $x\in B$, as you did, you should also specify that $x\notin A$. Clearly, if $x\in B$ and $x\in A$, then $x\notin (B\,\cap\,A^c)$.