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

Question

This is an exercise from Velleman's "How To Prove It". I strugled with this problem for a while, so I just want to make sure that it is correct.

Suppose $B \subseteq A$, and define a relation $R$ on $\mathscr{P}(A)$ as follows: $$R = \{ (X,Y) \in \mathscr{P}(A) \times \mathscr{P}(A) | X \vartriangle Y \subseteq B\}$$ Prove that for every $X \in \mathscr{P}(A)$ there is exactly one $Y \in [X]_R$ such that $Y \cap B = \emptyset$.

Proof: Let $X \in \mathscr{P}(A)$ be arbitrary. Let $Y = X \setminus B$. We must show that $X \vartriangle Y \subseteq B$ and $Y \cap B = \emptyset$.

Let $y \in X \vartriangle Y$ be arbitrary. Then $y \in X \setminus (X \setminus B) \cup (X \setminus B) \setminus X$. But since $(X \setminus B) \setminus X = \emptyset$, $y\in X \setminus (X \setminus B)$. This means that $y \in X$ and $y \in X \rightarrow y \in B$. It immediately follows that $y \in B$. Since $y$ was arbitrary, $X \vartriangle Y \subseteq B$, so $Y \in [X]_R$. Now suppose that $Y \cap B \neq \emptyset$. Then we can choose a $y$ such that $y \in Y \cap B = (X\setminus B) \cap B$. Then $y \in X \setminus B$ and $y \in B$. But then we have $y \in B$ and $y \notin B$, which is a contradiction. Thus, $Y \cap B = \emptyset$.

To show that $Y$ is unique, let $Y' \in [X]_R$ be arbitrary such that $Y' \cap B = \emptyset$. So $X \vartriangle Y' \subseteq B$. Since $X \vartriangle Y' \subseteq B$ and $X \vartriangle Y \subseteq B$, $(X \vartriangle Y) \vartriangle (X \vartriangle Y') = (Y \vartriangle X) \vartriangle (X \vartriangle Y') = Y \vartriangle (X \vartriangle X) \vartriangle Y' = Y \vartriangle Y' \subseteq B$. Suppose that $Y \vartriangle Y' \neq \emptyset$. Then we can choose an $x$ such that $x \in Y \vartriangle Y'$. Then either $x \in Y\setminus Y'$ or $x \in Y' \setminus Y$. In either case, we would have $x \in Y \cap B$ or $x\in Y' \cap B$, which are contradctions. Thus, $Y \vartriangle Y' = \emptyset$. It immediately follows that $Y = Y'$. $\square$

Answer 1

Your proof is okay.

Alternative for second part:

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Suppose that $Y\in\left[X\right]_{R}$ and $Y\cap B=\varnothing$.

Then $Y-X\subseteq B$ and combining that fact with $Y\cap B=\varnothing$ we conclude that $Y-X=\varnothing$ or equivalently that $Y\subseteq X$.

Again combining with $Y\cap B=\varnothing$ we find that $Y\subseteq X-B$.

If this inclusion is proper then some $x$ must exist with $x\notin Y$, $x\in X$ and $x\notin B$.

This however contradicts that $X-Y\subseteq B$ so we conclude that the inclusion is not proper so that: $$Y=X-B$$