Equivalence Classes Partition a Set Proof

Thank you

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)?

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

Thank you for such a logical reasoning 👍👍

Thanks!!

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

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

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?

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

Very good explanation.

It was really helpful. Thanks!

Better than my prof

Best teacher ever! Thank you for your time

Thank you very much!

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