Prove equivalence ratio and find class of $B\subseteq A$ , $X \sim Y \Leftrightarrow X \cap B = Y \cap B$

Question

Prove equivalence ratio and find class of $B\subseteq A$ , $X \sim Y \Leftrightarrow X \cap B = Y \cap B$.

Well, I've proved really easily that it is reflexsive, symmetrical and transitive.

But I'm really struggling to find the equivalence class.

Find: Equivalence class of $\{1,2,3\}$, Given $A = \{1,2,3,4\} , B = \{1,2,4\}$.

I do know the fundamentals and how to find equivalence classes of 'easier' problems, but I really got confused on this one. Any help would be highly appreciated.

note: I started by saying $B \in P(A)$ and $a = \{1,2,3\}$, I want to find $[a]_{\sim}$

and I also do know that if $a \in P(A)$ then $[a]_{\sim} = \{ x \in P(A) | (a,x) \in \sim \}$ (And from here I'm not sure how to put 'all the pieces together' to solve it)

Answer 1

Correct me if I'm wrong:

Do you want to find the equivalence class of the set $C \subseteq A$ ,given that $A\supseteq B$ and 2 sets $X,Y$ are equivalent $ \Leftrightarrow X\cap B = Y\cap B $ ?

The answer would be $[C]=\{ D \in \mathcal{P}(A) : D\cap B = C \cap B \}$ so if $C=a$ then $ C\cap B = \{1,2 \} $ $[a]=\{ D\in \mathcal{P}(A) : D\cap \{1,2,4\} = \{1,2 \} \}= \{ a,\{1,2 \} \}$

Because the condition implies that 1 and 2 must be in $D$; if $4\in D$ then the condition doesn't hold so it leaves only those sets

EDIT: Just a comment, usually sets are in Capital Letters , unlike $a$ !

Answer 2

Hint. The equivalence relation is defined on $P(A)$. Now, just read the definition: if $X, Y \in P(A)$, $X \sim Y$ if and only if $X \cap B = Y \cap B$. In other words, the equivalence class of $X$ only depends on the intersection of $X$ with $B$. What are the possibilities for the set $X \cap B$?