Question

Problem 5. Equivalence classes really separate elements in A in a very strict way. Two different equivalence classes are disjoint, altogether they give A back. Working with the defi- nition of equivalence class (R.S,T) and the definition of equivalence class prove the mentioned properties. (a) If a,b € A, then a – b if and only if (a)rn [b]R=. (b) If a, b e A, then if [a]rn (br #0 the [a]R= [b]R (c) For all a € A, (a)r +0.

Answer 1

a)

Suppose that Ai and Aj are any two distinct equivalence classes of R

and since Ai and Aj are distinct,

then Ai$\neq$ Aj

and since Ai and Aj are equivalence classes of R, so there must exists elements a and b in A such that Ai=[a] and Aj=[b]

By Lemma 8..3.3., either or [a]=[b]

(9] U D

But since [a]$\neq$[b] because Ai$\neq$ Aj , hence

(9] U D

b)

Suppose that Ai and Aj are any two distinct equivalence classes of R

and since Ai and Aj are equivalence classes of R, so there must exists elements a and b in A such that Ai=[a] and Aj=[b]

By Lemma 8..3.3., either or [a]=[b]

(9] U D

But since $[a]\cap [b]\neq \phi$ is given

hence [a]=[b]

Please give thumbs up.