Equivalence Classes Partition a Set Proof 

Best teacher ever! Thank you for your time

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

Thank you for such a logical reasoning 👍👍

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)?

Thank you very much!

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

It was really helpful. Thanks!

Thank you

Very good explanation.

Thanks!!

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

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

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

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

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?

Better than my prof

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

14:12 how?