Stuck on a beginner proof writing: If $U = A \cup B$ and $A \cap B = \emptyset$ then $A = U \setminus B$.

Question

While I was trying to work on the proof, I realized I never ended up using the second given, $A \cap B = \emptyset$. Then I realized my logic could be off entirely, but I'm not entirely too sure what about it needs fixing.

I'll be tackling the first part since that doesn't really seem to make sense. Feel free to not focus on the wording, and just aid with the logic portion of it.

Hello future reader! If you encounter this ~1 week of after I posted it, that means I temporarily took it off because I do have to submit it and can be plagiarized (regardless of the difficulty of the proof- it is a full solution that is written in my style). It'll be up in ~1 week as I said, I'd definitely like to keep it for someone unless to learn basic set theory proofs in the future!

Answer 1

(i) "then let's assume $x \in U \setminus B$." No, you want to prove $x \in U \setminus B$ using the assumption $x \in A$. By definition of set difference, you want to prove that $x \in U$ and $x \notin B$.

The rest looks fine.

Answer 2

Your argument is not bad, but it's too wordy. One can make it shorter by first noting that

$$(A \cup B) - B = A - (A \cap B) \hspace{20px}(\star)$$

So, when $A \cap B = \emptyset$, we get $(A \cup B) - B = A$.

To prove $\star$, let's take $x \in (A \cup B) - B$, we can write down:

$x \in (A \cup B ) - B \iff (x \in A \text{ or }x\in B) \text{ and }x \not\in B \iff x \in A \text{ and } x \not\in B$.

The last one says that $x$ is in $A$, but it cannot be in $B$. This equivalently means that the elements of $B$ that are in $A$ are excluded. Since $B$ in general is not a subset of $A$ writing this down as $x \in A - B$ doesn't seem right but it can be equivalently written down as $x \in A-(A\cap B)$