Symmetric Difference

**santiagos11** · November 27th 2009, 06:43 AM

I have a problem which s related to this symmetric propety.
Show that there exist a unique set $N$ such that $(A \Delta N)=A$ for all $A$ . The set $N$ is obviously the empty set. But then how do I prove the uniqueness part of $N$ , i.e. if $(A \Delta N1)=A$ and $(A \Delta N2)=A$ , then $N1=N2$ = the empty set

**Plato** · November 27th 2009, 07:50 AM

Originally Posted by santiagos11

I have a problem which s related to this symmetric propety.
Show that there exist a unique set $N$ such that $(A \Delta N)=A$ for all $A$ . The set $N$ is obviously the empty set.

$A\Delta N = A\, \Rightarrow \,\left( {A\backslash N} \right) \cup (N\backslash A) = A\; \Rightarrow \;A \cap N^c \subseteq A\;\& \;N \cap A^c \subseteq A$
We know that $N = \left( {N \cap A} \right) \cup \left( {N \cap A^c } \right)$ .
But that along with the above means that $N \subseteq A$ . WHY?

Now suppose that $\left( {\exists p} \right)\left[ {p \in N} \right]$ this means that $p \in N \cap A\; \Rightarrow \;p \notin N\backslash A \wedge p \notin A\backslash N$ .
Is that a contradiction?

**santiagos11** · November 27th 2009, 09:28 AM

thank you. You are very good in your arguments.

**santiagos11** · November 29th 2009, 06:43 PM

I have yet another simmilar problem.
Suppose $A,B$ are sets. Show that there exist a unique set $C$ such that $(A \Delta C)=B$ . If $A,B$ are disjoint, then $C=A \cup B$ . But if not, then what would C have to be?

**santiagos11** · November 30th 2009, 04:51 PM

never mind. i got it

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