Quotient Construction of a Power Set, equivalence relations, and symmetric difference.

Question

Fixing $N \in P(X)$ (the power set of X), we say that $A,B \in P(X)$ $agree \ away \ from \ N$ if $A - N = B - N$.

I have to show that $A \sim A', B \sim B'$ implies that $(A \Delta B) \sim (A' \Delta B')$, where $A \Delta B$ is defined as $(A - B) \cup (B - A)$. I am unsure how to begin this. Any help is appreciated.

Answer 1

You can start by writing $$ \begin{align} (A \Delta B) - N&= ((A - B)\cup (B - A)) - N \\ &= ((A - B)-N) \cup ((B - A)-N) \\ &= ((A-N)-(B-N)) \cup ((B-N)-(A-N)) \end{align}$$ and similarly for $(A' \Delta B') - N$ (of course, you should check why each of the above steps is valid). Now use your assumption to conclude that these sets are identical.