Michael Hong: regarding Puzzle 31, I don't think it makes sense to say \\(x \sim_p y \leftrightarrow \left (x_p,y_p \right ) = S_x\\) because:

1. You haven't said what \\( x_p \\) and \\( y_p \\) are, and I didn't define them either.

2. Even if we knew what \\( x_p \\) and \\( y_p \\) were, \\( (x_p,y_p) \\) means the ordered pair consisting of these two points, and it's impossible for that ordered pair to equal \\( S_x \\). Maybe you meant the subset \\( \\{x_p , y_p\\} \\).

3. The set \\(S_x\\) consists of _all_ points in \\( X \\) that are equivalent to \\(x\\), so it's unlikely to consist of just two points.

Maybe you were trying to say

$$ x \sim_P y \textrm{ if and only if } y \in S_x $$

That's true. So is

$$ x \sim_P y \textrm{ if and only if } x \in S_y \textrm{ and } y \in S_x $$

and

$$ x \sim_P y \textrm{ if and only if } S_x = S_y $$

These are some statements that looks sort of like yours but are true.

To prove there's a one-to-one correspondence between equivalence relations and partitions, one natural strategy is to use Puzzles 29 and 30 and then prove

$$ P_{\sim_P} = P $$

and

$$ \sim_{P_\sim} = \sim .$$

In other words, the map sending a partition \\(P\\) to its equivalence relation \\(\sim_P\\) is the inverse of the map sending an equivalence relation \\(\sim\\) to its partition \\(P_\sim\\).