# How to complete the proof that $A^c\cup(A\setminus B)=(A\cap B)^c$?

I had to prove that: For all sets A and B, $A^c \cup (A \setminus B) = (A \cap B)^c$.

Below is what I did, but I'm kind of stuck at the time.

So I begin with proving $A^c \cup (A \setminus B) \subseteq (A \cap B)^c$.
Let $x \in A^c \cup (A \setminus B)$.
Note, by DeMorgan's law, $(A \cap B)^c$ = $A^c \cup B^c$.
Then, $x \in A^c$ or $x \in A$ and $x \notin B$.
If $x \in A^c$, then $x \in A^c \cup B^c$.
If $x \in (A \setminus B)$, then $x \in A$ and $x \in B^c$, thus $x \in A^c \cup B^c$.

Left is to prove $(A \cap B)^c \subseteq A^c \cup (A \setminus B)$.
Let $x \in A^c \cup B^c$.
Then, $x \in A^c$ or $x \in B^c$.
If $x \in A^c$ then $x \in A^c \cup (A \setminus B)$.
If $x \in B^c$ then ..?

I don't think you can now say that $x \in A \setminus B$, can you? I considered coming up with a counterexample as I couldn't figure out what I'm missing, but I can't seem to find one.

Any help is appreciated! Thank you.

HINT: Now consider the two cases, either $x\in A$ or it isn't. One of them you dealt with. What does the other give you?
Full answer: (After indication from the comments that the hint was understood) if $x\in B^c$ then either $x\in B^c$ and $x\in A^c$, in which case we already know that $x\in A^c\cup(A\setminus B)$, or $x\in B^c$ and $x\in A$, in which case $x\in A\setminus B$ and the proof is completed.
• If $x \in A$, obviously $x \in A \setminus B$. – Alexei Oct 21 '14 at 22:54
• If $x \notin A$, then $x \notin A \setminus B$. How do I get to this from $x \in B^c$? I believe I have to show that $x \notin A$ and $x \notin B$, but I can only show the latter. – Alexei Oct 21 '14 at 22:58
• First of all, $x\notin A$ can also be written as? Secondly, you have that $x\in A$ or $x\notin A$. Regardless to $x\in B^c$, you just do a case by case analysis. – Asaf Karagila Oct 21 '14 at 23:00
• $x \notin A$ can be written as $x \in A^c$. Sorry if my questions are trivial, but I don't see why we have to consider them separately (because of the brackets). I thought we have to consider two cases: (1): $x \in A^c$ and (2): $x \in A$ and $x \notin B$. – Alexei Oct 21 '14 at 23:04