1. ## Simple Set Question

To prove that there exists a unique set $N \in \mathcal{P}(X)$ such that $A \Delta N = A$ for all $A \in \mathcal{P}(X)$ is it as simple as choosing $N = \emptyset$ and so:

$(A \cup \emptyset) - (A \cap \emptyset) = A - \emptyset = A$?

2. I guess its right then?

Thanks

3. Not quite. You have shown existence but not uniqueness.

4. To show uniqueness: We want to find a set $N \in \mathcal{P}(X)$ such that $(A \cup N) - (A \cap N) = A$. This means that $A \cup N = A$ and $A \cap N = \emptyset$ so that $A - \emptyset = A$.

By definition, $A \cup N = A \Rightarrow N = \emptyset$ (which is unique). And so it follows that $A \cap \emptyset = \emptyset$ and $N = \emptyset$.

Would this show uniqueness?

5. This what must be done.
$A\Delta B = A\quad \Rightarrow \quad B = \emptyset$

6. Suppose that $x \in A \Delta B$ and $x \in A$. Then $x \in (A \cup B) - (A \cap B) \Rightarrow x \in A$ and $x \in A \Rightarrow x \in (A \cup B) - (A \cap B)$. It follows that $B = \emptyset$.

Is this correct?

7. No that does not work.

I think it is safe to say that most of us use this equivalence:
$A\Delta B = \left( {A - B} \right) \cup \left( {B - A} \right)$.

With that we can say if $A\Delta B = A$ then that means $\left( {A - B} \right) \cup \left( {B - A} \right) = A$.
From which we can conclude that $\left( {B - A} \right) \subseteq A\quad \Rightarrow \quad B = \emptyset$.

QED

8. Why doesn't it work?

$x \in A \Rightarrow x \in A$?

Oh ok: Its apparent that if $B-A \subseteq A$ then $B = \emptyset$. Sort of like a contradiction if we didn't have the empty set.