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\$$.