# If symmetric difference leaves a set unaffected, the second set is empty.

To prove:

If $A \triangle B=A$, then $B= \emptyset$.

This seems simple enough, as an idea. I mean if the set $B$ is anything but empty, $A\triangle B$ would contain more or less than simply $A$, simple examples would be two disjoint sets or two intersecting sets or whenever $A=B$. But how do I prove it formally?

The example I gave feels like I can make it work with a contrapositive: If $B$ is not empty, $A\triangle B$ does not have the same elements as $A$.

• You should know that $A \Delta A = \emptyset$. Then $$A \Delta A \Delta B =A \Delta A$$ $$\emptyset \Delta B= \emptyset$$ $$B= \emptyset$$ – Crostul Sep 5 '18 at 16:11
• That's brilliant. – Wesley Strik Sep 5 '18 at 18:14
• Nice for a direct proof :) – Wesley Strik Sep 5 '18 at 18:24

Let us recall that the symmetric difference is defined as $$A \triangle B = (A\setminus B)\cup(B\setminus A)$$
Thus, if $B$ was not empty, it would contain an element $b$. I distinguish two cases:
1. If $b$ also lies in $A$, then by hypothesis it lies in $A \triangle B=A$. This is absurd, because $A \triangle B$ does not contain any element of $A\cap B$.
2. If $b$ does not lie in $A$, then it lies in $B\setminus A$. Hence, by definition, it must lie in $A \triangle B$. But by hypothesis, this is no other than $A$, which leads us to an absurdity.
We conclude that $B$ must be empty.