1. Each \\(A_p\\) is nonempty because it is \\((\sim)\\)-connected, and a \\((\sim)\\)-connected subset is nonempty by definition.

2. If \\(A_p\\) and \\(A_q\\) had a nonempty intersection, they would each contain some element \\(x \in X\\). But then, since they are \\((\sim)\\)-closed and \\((\sim)\\)-connected, \\(A_p\\) and \\(A_q\\) would each contain *all and only* elements of X equivalent to x, contradicting that \\(A_p\\) and \\(A_q\\) are unequal. So \\(A_p\\) and \\(A_q\\) canâ€™t have a nonempty intersection.

3. We prove that A and the specified union are subsets of each other: First, every element in the union is obviously in A, because the union is a union of *subsets* of A. On the other hand, if \\(x \in A\\), then the set \\(A_r\\) of all elements \\(y \in A\\) with y \\(\sim\\) x is \\((\sim)\\)-closed and \\((\sim)\\)-connected (by transitivity and reflexivity of equivalence). So \\(A_r\\) is by definition one of the subsets in the specified union, and thus x (which is in \\(A_r\\) by reflexivity) is in that union.

2. If \\(A_p\\) and \\(A_q\\) had a nonempty intersection, they would each contain some element \\(x \in X\\). But then, since they are \\((\sim)\\)-closed and \\((\sim)\\)-connected, \\(A_p\\) and \\(A_q\\) would each contain *all and only* elements of X equivalent to x, contradicting that \\(A_p\\) and \\(A_q\\) are unequal. So \\(A_p\\) and \\(A_q\\) canâ€™t have a nonempty intersection.

3. We prove that A and the specified union are subsets of each other: First, every element in the union is obviously in A, because the union is a union of *subsets* of A. On the other hand, if \\(x \in A\\), then the set \\(A_r\\) of all elements \\(y \in A\\) with y \\(\sim\\) x is \\((\sim)\\)-closed and \\((\sim)\\)-connected (by transitivity and reflexivity of equivalence). So \\(A_r\\) is by definition one of the subsets in the specified union, and thus x (which is in \\(A_r\\) by reflexivity) is in that union.