Equivalence Classes Partition a Set Proof

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-envb]

Best teacher ever! Thank you for your time

[DP-sqlw]

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)?

[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?

[rampagex]

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

[Sky-nthy]

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

[Sky-nthy]

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?

[SiddharthBaruaGeoGeek]

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

[mr.anonymous]

Thank you for such a logical reasoning 👍👍

[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

[sophieliu]

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

[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 ?