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> a Subset cannot have greater cardinality than the set it derives from. I am not completely certain in the subtleties of non-standard set theory, but at least

> a Subset cannot have greater cardinality than the set it derives from. I am not completely certain in the subtleties of non-standard set theory, but at least if we replace the axiom of Choice (AC) by the axiom of Determinacy (AD) - which is perfectly fine, though it is less common to do mathematics in this system - we can partition the real line R by an equivalence relation that has more equivalence classes than elements of R. So at least we can partition R into more disjoint subsets than R has elements. Source: https://math.stackexchange.com/questions/1104028/what-are-disasters-with-axiom-of-determinacy/1148609#1148609 https://math.stackexchange.com/questions/1104028/what-are-di...

xelxebar 9y ago
Oh cool. Thanks for sharing this!
titzer 9y ago
> 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.
- wolfgke 9y ago
  > 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.

xelxebar 9y ago

Oh cool. Thanks for sharing this!

titzer 9y ago

> 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.

wolfgke 9y ago
> 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.