Finish proof Proposition 1.19. Suppose that \(\sim\) is an equivalence relation on a set A, and let P be the set of \(\sim\)-closed and \(\sim\)-connected subsets.
Show that each part \(A_p\) is nonempty
Show that \(p \ne q \implies A_p \cap A_q = \varnothing\)
Show that \(A = \bigcup_{p \in P} A_p\)
Part of the definition of \(\sim\)-connected is being nonempty
Suppose \(a \in A\) is in the intersection. Then \(a \sim p\) and \(a \sim q\) for some elements \(p \not\sim q\) arbitrarily selected from \(A_p, A_q\). But this is impossible because \(\sim\) is transitive, so this must be impossible.
Every \(a \in A\) is part of some equivalence class which is a \(\sim\)-closed and \(\sim\)-connected set, so \(A \subseteq \bigcup_{p \in P} A_p\)
The equivalence class is \(\sim\)-closed because two elements being \(\sim\)-related implies they are in the same equivalence class.
The equivalence class is \(\sim\)-connected because equivalence classes are nonempty and the equivalence relation is transitive.
The constituents of \(A_p\) were defined to be subsets of \(A\), so unioning these will also be a subset of \(A\), i.e. \(\bigcup_{p \in P} A_p \subseteq A\).
Therefore \(A = \bigcup_{p \in P} A_p\).