Equivalence Classes Partition a Set Proof

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Equivalence Classes Partition a Set Proof. This video starts with the definition of an equivalence class and then proves that for a given set S and an equivalence relation R on S, we can write S as the disjoint union of it's equivalence classes. Every single detail is shown or explained in the proofs.

[@MrCoreyTexas]

Thanks for this material, I kept hearing about equivalence classes, and I forgot what they were. I find that the more abstract "abstract algebra" is, the harder time I have keeping up :)

[@Christian-en7vb]

Best teacher ever! Thank you for your time

[@TheMathSorcerer]

😄

[@TheMathSorcerer]

[@rampage14x13]

Thank you! I am self-studying Munkres's Topology atm and this cleared up my doubts!

[@TheMathSorcerer]

Glad it was helpful!

[@mr.anonymous179]

Thank you for such a logical reasoning 👍👍

[@DP-sq7lw]

This is so beautiful!!!!! Thank you!!!

[@DP-sq7lw]

For the second part of the last proof, how do we know that C_i is not equal to C_j if i is not equal to j? The proposition only shows that given R is equivalence relation, the equivalence classes are either disjoint or equal to each other. Is it possible that the equivalence classes for different elements in set A are the same (i not equal to j, but C_i = C_j)?

[@SiddharthBaruaGeoGeek]

Really Well explained! you seem to be a Math Lecturer..

[@andrewsharon4066]

Very good explanation.

[@TheMathSorcerer]

Thank you👍

[@MrCoreyTexas]

I'll have to think about how the set of equivalence classes partitions a set vs. how cosets also partition a set. Differences and similarities?

[@TheAllen501]

Better than my prof

[@TheMathSorcerer]

Thank you

[@김용준-o5z]

It was really helpful. Thanks!

[@TheMathSorcerer]

You're welcome!

[@Sky-nt1hy]

Another definition i saw was [s]={x| (s,x)element R} But isnt it sRx not xRs? Hope you got what i mean. Im so confused

[@JesusHernandez-xv7lf]

[s]={x|(s,x) element R} ={x|sRx}. It's the same definition of relation

[@JesusHernandez-xv7lf]

If R is an equivalence relation, then by reflexivity xRs=sRx

[@TheMathSorcerer]

Ya alternate def same thing so yeah

[@milkman2808]

@@JesusHernandez-xv7lf *by symmetry

[@asparagii]

Hello - thanks. A dumb question for someone who knows more about this than I do: at 5:30 you start a proof by contradiction of [s] intersection [t] = empty set. I understand contradiction starts with supposing the negation of a statement - but i thought intersection was equivalent to 'and' and so when applying DeMorgan to the statement, you'd get "[s] or [t] does not = empty set." (Alternatively, replace 'or' with union). I'm guessing I'm wrong after having watched this - but could someone please explain why? Thanks

[@TheMathSorcerer]

ok the statement is, "S intersect T = empty set", so if this is NOT true, then this means "S intersect T != empty set" And then this means there is an element in the intersection. We didn't apply DeMorgans

[@asparagii]

@@TheMathSorcerer ok thanks very much for the quick response Professor. I get that you make the statement false by changing equals to not equals (and it makes sense that "the intersection contains nothing", versus "the intersection contains something" are opposites). I just thought that negating something was the same as making it false, and I thought you'd have to apply DeMorgan to do that (and then distribute the 'not' through the statement, which would change intersects to union).

[@Sky-nt1hy]

For the beginning, How come it is xRs? ex, S={1,2,3} R={(1,2)} [1]={2} right? Then but by definition of the video it says 2R1 but there is no ordered pair (2,1). Shoudn’t it be sRx instead of xRs?

[@avraham25]

I know it's already late but whatever, maybe it'll still help. In your example R isn't equivalent relation, since R needs to be reflexive, symmetric and transitive. In that case you'll see that for every a,b in R if aRb then bRa and that's why [a]=[b].

[@Carrarvella]

Thank you

[@sophieliu8419]

For the second case where statement 2 holds, can you do without loss of generality?

[@brunojimenez6086]

Thanks!!

[@AraDeanMaffy]

at 13:46 isnt xE[x]=c exactly what we want to show ? so we cant use it for the proof ? can someone elaborate ?

[@mwaleed6249]

x€S and xRx always => x€ [x]

[@mwaleed6249]

x€S and xRx always => x€ [x]

[@1997tennis]

Thank you very much!

[@a.nelprober4971]

14:12 how?

[@a.nelprober4971]

Do we replace the ci with s

[@andyalbertochavez5135]

@@a.nelprober4971 Not sure what you're asking but it's saying if x is an element in S, then it's an element in [x], which is one of the equivalence classes (c_i), so it's in the union. As a result, any element in S is also in the union so S is a subset of that union.

[@a.nelprober4971]

@@andyalbertochavez5135 thanks bro. I'm so shit at discrete