How to show that if $A\mathbin{\triangle}C=B\mathbin{\triangle}C$ for all $C$, then $A=B$? I have no idea how to start the proof.
$\triangle$ means the symmetric difference.
Prove or Disprove the following:
For every 2 sets $A$ and $B$
if for every $C$
$A \mathbin{\triangle} C = B \mathbin{\triangle} C $
then 
$A=B$
 A: Recall that the set $A \mathbin{\triangle} C$ is defined to be the set of all elements that belong to either $A$ or $C$ but not both.
Notice that if we take $C$ to be the empty set then $A \mathbin{\triangle} C=A$.
Therefore, you can prove the statement by just taking $C$ to be the empty set.
It might be worth noting that a more general result is true. That is, if there exists some set $C$ such that $$A\mathbin{\triangle} C = B\mathbin{\triangle} C$$ then $$A=B$$
Proof: Suppose $a\in A$. There are two cases to consider.
Case 1: If $a\in C$, then $a\notin A \mathbin{\triangle} C$ since $a$ lies in both $A$ and $C$. So $a\notin B \mathbin{\triangle} C$, whence $a\in B$.
Case 2: If $a\notin C$, then $a\in A \mathbin{\triangle} C$ since $a$ lies in $A$ but not $C$. So $a\in B \mathbin{\triangle} C$, whence $a\in B$.
So we have shown that any element of $A$ lies in $B$. Similarly, one can show that any element of $B$ lies in $A$. Therefore, $A=B$. $\square$
A: The point is that if you are given the assumption that "For every $x$ such and such happens", you are free to place whatever $x$ into the equation.
This is very similar, and in fact a specific case,1 of the statement that if $\Bbb F$ is a field, and for every $c$, $a+c=b+c$, then $a=b$. How would you have solved that? You would have argued that since this equation is true for every $c$, in particular it is true for $c=0$, and in that case we get $a=a+0=b+0=b$, so $a=b$.
Here we have a similar situation, where taking $C=\varnothing$ solves the problem in a similar fashion. You are, however, expected to argue why $A\mathbin{\triangle}\varnothing=A$, for an arbitrary set $A$.



*

*This is a specific case, since $\mathcal P(X)$ along with $\triangle$ and $\cap$ form a field where $\triangle$ gives us $+$ and $\cap$ gives $\cdot$; so $A\mathbin{\triangle}\varnothing$ is the same as $A+0$.

