Paulius: the problem says

> Let \\(P\\) be the set of \\((\sim)\\)-closed and \\((\sim)\\)-connected subsets \\((A_p)_{p \in P}\\).

That's a bit of a mouthful, but the idea is this. If we start with an equivalence relation \\(\sim\\) on a set \\(X\\), we can use it to chop \\(X\\) into a bunch of parts called 'equivalence' classes:

\\(P\\) is the set of these parts. It's a bit ridiculous to say "the set of subsets \\((A_p)_{p \in P}\\)". One can equivalently just say \\(P\\), since this is the same thing!

If you're wondering what a " \\((\sim)\\)-closed and \\((\sim)\\)-connected subset" is, you can ask me or look it up in the book - it's in the proof of Proposition 1.11, right before this exercise.

The point of the exercise is to show that these "\\((\sim)\\)-closed and \\((\sim)\\)-connected subsets" really do form a partition of \\(X\\).

> Let \\(P\\) be the set of \\((\sim)\\)-closed and \\((\sim)\\)-connected subsets \\((A_p)_{p \in P}\\).

That's a bit of a mouthful, but the idea is this. If we start with an equivalence relation \\(\sim\\) on a set \\(X\\), we can use it to chop \\(X\\) into a bunch of parts called 'equivalence' classes:

\\(P\\) is the set of these parts. It's a bit ridiculous to say "the set of subsets \\((A_p)_{p \in P}\\)". One can equivalently just say \\(P\\), since this is the same thing!

If you're wondering what a " \\((\sim)\\)-closed and \\((\sim)\\)-connected subset" is, you can ask me or look it up in the book - it's in the proof of Proposition 1.11, right before this exercise.

The point of the exercise is to show that these "\\((\sim)\\)-closed and \\((\sim)\\)-connected subsets" really do form a partition of \\(X\\).