>>13740655Since the { Ai } partition A, then by definition:
Union(Ai) = A (for i=1,n)
Intersect(Ai, Aj) = {} (for i != j).
Therefore, we can conclude that if a in A, then a in Ai for some Ai and only in Ai (and not some other Aj).
Well-defined: Trivial, since set membership is also well-defined.
a ~ a: Trivial, since a in A => a in Ai and if a in Ai, then a in Ai tautologically. Thus, a ~ a.
a ~ b => b ~ a: Also trivial since a ~ b => a, b in Ai for some Ai, so b, a are also in some Aj (in this case j = i).
a ~ b AND b ~ c => a ~ c: We know a, b in Ai and b, c in Aj. We want to show a, c in Ak. Note that since { Ai } partitions A, if b in Ai, then b cannot be in any other Aj for j != i. Thus, Aj = Ai.
Therefore, a, b in Ai and b, c in Ai trivially implies a, c in Ai (so Ak = Ai). Thus this property also holds.
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Also, keep in mind that (infinite) sets in modern mathematics are not well-defined and Cantor was a literal schizo. That is all.
