How to prove if $(A \cup B) - (A \cap B) = A$ is true, then $B = \emptyset$?

Question

I have figured out that this is true, but I'm not quite able to prove it.

I have tried direct proof, contrapositive and contradiction.

For contradiction, I assumed that B is not the empty set.

Then, I wrote that the difference between A and the empty set is the only operation that equals back A. (For the premise) $(A-\emptyset = A)$

Then $A\cap B$ only equals empty set when B is the empty set. Thus B has to be the empty set, therefore contradicting and proving that B is the empty set.

Does this make sense?

I also tried direct proof, where I simplified the premise to $(\overline A \cap B) \cup (A\cap \overline B)$ through set identities, but this doesn't seem to help me determine that $ B = \emptyset$

Any help is appreciated, thanks!

Answer 1

Proof. By contradiction. Suppose $B\ne \varnothing$. Then there exists some element $x\in B$. Now, we have two possible cases:

Case 1. $x\in A$. In this case, $x\in A\cup B$ and $x\in A\cap B$, then $x\notin (A\cup B)-(A\cap B)=A$, a contradiction to $x\in A$.

Case 2. $x\notin A$. In this case, $ x\in A\cup B$ and $x\notin A\cap B$, then $x\in (A\cup B)- (A\cap B)=A$, a contradiction to $x\notin A$.

Since we have a contradiction in both cases, we conclude $B=\varnothing.$

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Also, $\varnothing$ is not the only set such that $A-B=A$. Any disjoint set (that is $A\cap B=\emptyset$ will check $A-B=A$.

Answer 2

Suppose $x \in B$. Then $x \in A \cup B$.

If $x \in A$, then $x \in (A \cup B) - (A \cap B)$: by definition of difference, $x \notin A \cap B$. This contradicts ($x \in B$ and $x \in A$).

If $x \notin A$, then $x \notin A \cap B$. By definition of difference, this means that $x \in (A \cup B) - (A \cap B)$. But this set is $A$, so you have a contradiction.

In other words, for all $x \in B$ you have a contradiction. This means that $B$ is empty.

Answer 3

Since $(A\cup B)-(A\cap B)=(A\setminus B)\cup(B\setminus A)$ and $A=(A\setminus B)\cup(A\cap B)$, and these are each unions of disjoint sets, $B\setminus A=A\cap B$. So each $x\in B$ is in $A$ iff it isn't, i.e. $B$ is empty.

Answer 4

This is simple algebra if you know that $(A\cup B)-(A\cap B)$ is the symmetric difference and that the symmetric difference is associative: from $A\mathbin{\triangle}B=A$ we deduce $$ A\mathbin{\triangle}(A\mathbin{\triangle}B)=A\mathbin{\triangle}A=\emptyset $$ and therefore $$ \emptyset=(A\mathbin{\triangle}A)\mathbin{\triangle}B=\emptyset\mathbin{\triangle}B=B $$

Alternative proof.

Suppose $x\in A$; then $x\in (A\cup B)-(A\cap B)$ implies $x\notin A\cap B$, so $x\notin B$.

Suppose $x\notin A$. Then $x\notin(A\cup B)-(A\cap B)$, so either $x\notin A\cup B$ or $x\in A\cap B$. The latter case is not possible, hence $x\notin A\cup B$, in particular $x\notin B$.

In either case, $x\notin B$. Therefore $B=\emptyset$.