> we can partition the real line R by an equivalence relation that has more equivalence classes than elements of R
That's not surprising, since an equivalence relation is a function of two elements of the set, i.e. O(N^2), and a subset is a function of a single element, i.e. O(N) . The OP mentioned subsets, and an equivalence relation is not a subset of the original set--it doesn't even typecheck.
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That's not surprising, since an equivalence relation is a function of two elements of the set, i.e. O(N^2), and a subset is a function of a single element, i.e. O(N) . The OP mentioned subsets, and an equivalence relation is not a subset of the original set--it doesn't even typecheck.
I don't understand your reasoning. Every equivalence relation partitions the set S on which it is defined on into subsets - these are called equivalence classes. Every equivalence class contains at least one element of S. So I find it quite surprising that there are more equivalence classes than elements of S (in this case S are the real numbers) if we assume AD.
That's not surprising, since an equivalence relation is a function of two elements of the set, i.e. O(N^2), and a subset is a function of a single element, i.e. O(N) . The OP mentioned subsets, and an equivalence relation is not a subset of the original set--it doesn't even typecheck.