Proving $Y$ such that $Y \cap B = \emptyset$

Question

I have been solving this problem from Velleman's How to prove book:

Suppose $B \subseteq A$ and define a relation $R$ on $\mathcal{P}(A)$ as follows:

$R = \{(X,Y) \in \mathcal{P}(A) \times \mathcal{P}(A) \mid (X \Delta Y) \subseteq B\}$

a) Prove that $R$ is an equivalence relation on $\mathcal{P}(A)$.

b) Prove that for every $X \in \mathcal{P}(A)$ there is exactly one $Y \in [X]_R$ such that $Y \cap B = \emptyset$

Now I have proved the first part of the question. But I'm stuck in the second part. I cannot find any existential example for $Y$ for which $Y \cap B = \emptyset$. Any pointers on how to solve it ?

Answer 1

First try proving the statement for $X \in \mathcal{P}(B)$, where there's a fairly obvious candidate for $Y$. Then see if you can generalize.

Answer 2

Note that if $X\Delta Y\subseteq B,$ then both $X\setminus Y$ and $Y\setminus X$ must be subsets of $B.$ If $Y$ is disjoint from $B,$ it follows that $Y\setminus X=\emptyset,$ which happens if and only if $Y\subseteq X.$

Hence, we need a subset $Y$ of $X$ such that $X\setminus Y\subseteq B$ and $Y\cap B=\emptyset.$ Can you take it from there? (Try playing with some examples on Venn diagrams, say with $A$ a particular small finite set, and with $X$ and $B$ as subsets of $A.$ Once you've set up the Venn diagram, see if you can figure out what $Y$ is, in terms of $X$ and $B$. Once you've got the idea, try to generalize the argument. If you're still stuck, feel free to let me know what you're having trouble with, and I'll see what I can do to extricate you.)