Important Math Proof: The Set of Equivalence Classes Partition a Set

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In this video I prove a very important result in mathematics. Given an equivalence relation R on a nonempty set A, the set S of equivalence classes of A is a partition of A. Stated another way, this result says we can write A as a disjoint union of equivalence classes.
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19 июл 2022

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Комментарии : 18

@elcubanoz Год назад

Props for approaching a subject that is beyond the normal exposure to mathematics of a tipical highschool or even college graduate! And good choice of subect in terms of difficulty. In case you want some unsolicited advise: you mentioned this result is cool, but didn't really give a example of something cool you can do with it. It would be nice to have an example of the result being used

@jingyiwang5113 11 месяцев назад

Thank you so much for your detailed explanation! I have finally understood it.

@valeriereid2337 Год назад

Thank you so very much for explaining this. Your video most certainly helped me to understand this better.

@lingzhao5719 Год назад

For symmetry we can say if xRa then aRx and see if it satisfies.

@BeeChasnyAgario 10 месяцев назад

Very nice explanation!

@lingzhao5719 Год назад

Transitivity is more obvious for me as a b c are elements of A then if aRb and bRc then aRc and see if it satisfies

@alastairbowie Год назад

chill vids. thanks for the uploads (=

@Sanchuniathon384 Год назад

That's a really good way of looking at the Partition Problem. A lot of solutions rely on brute force, dynamic programming, etc. and in this case you have a way to do this declaratively by inspecting each element of the set. Heck, if you know a little bit about the problem in advance, you can even specify the equivalence classes, making the partition operation more computationally feasible.

@lingzhao5719 Год назад

R is a subset of AxA?

@TheMathSorcerer Год назад

yeah

@lingzhao5719 Год назад

@@TheMathSorcerer if I give you a math problems do you think you'll be able to solve it? It's about Cauchy Schwartz series

@wiggles7976 Год назад

I wanted to explicitly show that x belonged to a unique equivalence class using the definition of unique existence, just because of the idea of a widespread trend of computers checking proofs, which might get popular in the near future. We want to show that Forall x, Exists![y](x in [y]), or in other words, that there's a unique equivalence class [y] of x. The definition of unique existence is that there exists an element m such that P(m), and that if any other element n is such that P(n), then m=n. We let x be an arbitrary element of A. We showed x in [x]. Then we let [b] be an arbitrary equivalence class that is a subset of A. We showed that if x in [b], that [x]=[b]. (You used [a] instead of [x], but of course, your logic shows just the same that if x in [x] and x in [b], then [x]=[b]). Since [b] was arbitrary, it is the case that (changing the dummy variable from [b] to [z]) Forall [z], x in [z] implies [x]=[z]. Combining statements, we have that x in [x] AND Forall [z], x in [z] implies [x]=[z]. I'm not to sure on how the scope of quantifiers works if I want to do an existential generalization on [x], but I believe it would go like this seeing as how [x] is free in the "Forall [z], x in [z] implies [x]=[z]" statement: Exists[y](x in [y] AND Forall [z], x in [z] implies [y]=[z]). Then since x is arbitrary, we had a universal derivation of the statement Forall x, Exists[y](x in [y] AND Forall z, x in [z] implies [y]=[z]). In more abstract writing, just focusing on the existence part of the statement, that would say Exists Y (P(Y) AND Forall Z (P(Z) implies Y=Z). I think that checks out?

@BikramKumar-uy8nl Год назад

Nice recommendation RU-vid. I don't understand this high level maths. :(

@BikramKumar-uy8nl Год назад

I commented without watching the video. An example of this can be modulo function on set of integers. All numbers belong to a specific class, but a number does not belong in 2 different classes, thus partitioning it. I have some trouble with math language. But examples make it easier for me..

@littlemisterlong7292 Год назад

so where is antisymmetric?

@user-tv8qg5sq8d Год назад

We don't need to prove that all the elements in equivalence relation set form the whole set A?

@fonso-ir8xz 4 месяца назад

I initially thought that too. But after pondering a while, I think that by showing that every element of A belongs to exactly one equivalence class we show that all elements of equivalence relation set from the whole set A