Fixing $N \in P(X)$ (the power set of X), we say that $A,B \in P(X)$ agree away from $N$ if $A - N = B - N$. We denote $A \sim B$ if $A - N = B - N$.
I have to show that every equivalence class has a unique element which also belongs to $P(X - N)$ (that is, exactly one element of the class is a subset of $X - N$). I am unsure how to begin this. Any help is appreciated.
There are two parts to this problem.
Existence.
First try an example. Let $X=\{1, 2, 3, 4\}$ and $N=\{1, 2\}$. Take $A=\{1, 3, 4\}$; can you find some $B$ which agrees with $A$ away from $N$, such that $B\in P(X-N)$? NOTE: saying "$B\in P(X-N)$" is the same as saying "$B\subseteq X-N$."
If you can find such a $B$, think about how you came up with it; now try to generalize that.
Uniqueness.
Suppose $B$ and $B'$ agree away from $N$, and $B, B'\in P(X-N)$. What can you say about $B$ and $B'$? Why is this relevant?