Daniel@30: A typical way to prove \\(A = \cup_{p\in P}{A_p} \\) is by double inclusion. Since \\(A_p\\) are all subsets of \\(A\\) one inclusion is clear. For the other, suppose there is an \\(a\in A\\) and \\(a\not\in\cup_{p\in P}{A_p} \\), what can be said about \\(\{a\}\\) ?

Edit: For some reason the brackets are not showing, both "a" are supposed to be the set containing only "a".

SPOILER: \\(\{a\}\\) is a \\(\sim\\)-connected and \\(\sim\\)-closed subset, contradiction.

Edit: For some reason the brackets are not showing, both "a" are supposed to be the set containing only "a".

SPOILER: \\(\{a\}\\) is a \\(\sim\\)-connected and \\(\sim\\)-closed subset, contradiction.