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

Question

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

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

My attempt:

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Since $R$ is an equivalence relation, I use the notation:

$[X]_R = \{Y \in \mathscr P(A) \mid (Y,X) \in R \}$

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Suppose there are $T,Q \in [X]_R$, such that $T \cap B = \emptyset$ and $Q \cap B = \emptyset$. Need to prove that $T = Q$.

Take $a \in T$. Note that $a \notin B$.

We have $(T,X) \in R$. Since $T \Delta X \subseteq B$, it follows that $a \in X$.

Since $Q \in [X]_R$, it follows that $(Q,X) \in R$. Since $a \in X$ and $a \notin B$, it follows that $a \in Q$. Arbitrary element was considered, hence $T \subseteq Q$.

I will omit proof of $ Q \subseteq T$, because it is almost identical. $\Box$

Is it correct?

Answer 1

Your proof is correct but not complete. Let me explain what I mean.

A statement of the form "there exists exactly one $Y$ such that..." claims two things simultaneously:

1. there exists at least one $Y$ such that...;
2. there exists at most one $Y$ such that... .

What you have written correctly proves only point 2 (indeed, you proved that if $T$ and $Q$ are such that... then $T= Q$), but the proof of point 1 is missing. You can find it below.

We want to prove that for every $ \in \mathscr{P} ()$ there is at least one $ \in []_$ such that $ \cap = \emptyset$. Let $ \in \mathscr{P} ()$ and let $Y = X \smallsetminus B$. Then,

• $Y \in [X]_R$, indeed $X \, \Delta \,(X \smallsetminus B) \subseteq B$ (actually you can easily prove that $X \, \Delta \,(X \smallsetminus B) = X \cap B$);

• $Y \cap B = \emptyset$ because $(X \smallsetminus B) \cap B = \emptyset$ (indeed if $x \in X \smallsetminus B$ then $x \notin B$).