For any two sets, $A - B = B - A$ implies $A = B$

Question

Is the following statement True or False:

For any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$.

If it is true, prove it, otherwise provide a counterexample.

I am unable to come up with a counter example. I think the statement is true but how do I prove it?

Answer 1

If $A-B=B-A $ then for any $x\in A-B=B-A $ we $x\in A;x\in B; x\not \in A; x\not \in B $. That's a contradiction so $A-B=B-A $ is empty.

Thus there are no elements in $A $ that are not in $B$. In other words $A $ is a subset of $B $. Likewise there are no elements of $B $ that are in $A $. So $B $ is a subset of $A $.

So $A=B $.

Answer 2

If $A \setminus B = B \setminus A$, then

$A=A \setminus B \cup (A\cap B)= B \setminus A \cup (B \cap A) = B$.

Answer 3

Let’s use some Boolean algebra, in order to show a different point of view.

Let $C=A\cup B$; for a subset $X$ of $C$, denote $X^c=C\setminus X$; thus $$ A\setminus B=A\cap B^c,\qquad B\setminus A=B\cap A^c=A^c\cap B $$ Then \begin{align} A&=A\cap C && \text{because $A\subseteq C$} \\ &=A\cap (B\cup B^c) && \text{because $C=B\cup B^c$} \\ &=(A\cap B)\cup(A\cap B^c) && \text{distributivity} \\ &=(A\cap B)\cup(A^c\cap B) && \text{hypothesis} \\ &=(A\cup A^c)\cap B && \text{distributivity} \\ &=C\cap B && \text{because $A\cup A^c=C$} \\ &=B && \text{because $B\subseteq C$} \end{align}

You also have \begin{align} A\cap B^c &=(A\cap B^c)\cap(B\cap A^c) && \text{hypothesis} \\ &=A\cap(B^c\cap(B\cap A^c)) && \text{associativity} \\ &=A\cap((B^c\cap B)\cap A^c) && \text{associativity} \\ &=A\cap(\emptyset\cap A^c) && \text{because $B\cap B^c=\emptyset$} \\ &=A\cap\emptyset \\ &=\emptyset \end{align}