Symmetric Difference

Printable View

• Nov 27th 2009, 06:43 AM
santiagos11
Symmetric Difference
I have a problem which s related to this symmetric propety.
Show that there exist a unique set $\displaystyle N$ such that $\displaystyle (A \Delta N)=A$ for all $\displaystyle A$. The set $\displaystyle N$ is obviously the empty set. But then how do I prove the uniqueness part of $\displaystyle N$, i.e. if $\displaystyle (A \Delta N1)=A$ and $\displaystyle (A \Delta N2)=A$, then $\displaystyle N1=N2$= the empty set
• Nov 27th 2009, 07:50 AM
Plato
Quote:

Originally Posted by santiagos11
I have a problem which s related to this symmetric propety.
Show that there exist a unique set $\displaystyle N$ such that $\displaystyle (A \Delta N)=A$ for all $\displaystyle A$. The set $\displaystyle N$ is obviously the empty set.

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

Now suppose that $\displaystyle \left( {\exists p} \right)\left[ {p \in N} \right]$ this means that $\displaystyle p \in N \cap A\; \Rightarrow \;p \notin N\backslash A \wedge p \notin A\backslash N$.
Is that a contradiction?
• Nov 27th 2009, 09:28 AM
santiagos11
thank you. You are very good in your arguments.
• Nov 29th 2009, 06:43 PM
santiagos11
I have yet another simmilar problem.
Suppose $\displaystyle A,B$ are sets. Show that there exist a unique set $\displaystyle C$ such that $\displaystyle (A \Delta C)=B$. If $\displaystyle A,B$ are disjoint, then $\displaystyle C=A \cup B$. But if not, then what would C have to be?
• Nov 30th 2009, 04:51 PM
santiagos11
never mind. i got it