Let A be a non-empty set, and p an equivalence relation on A . Let a , b be an element of A . Prove that [ a ] = [ b ] is equivalent to apb

Question

the question: a) Let A be a non-empty set, and p an equivalence relation on A . Let a , b be an element of A . Prove that [ a ] = [ b ] is equivalent to $apb$ b) If p is both an equivalence relation and (simultaneously) a partial order on A , describe p.

this is currently what I have done for question a.

there exists $x$ element of [b] intersect [b]

therefore xRa and xRb

therefore aRx by symmetry rule

therefore aRb by transitive rule

therefore by lemma 1 (which wass a proof that "if R is an equivalence relation on a set A, and if a,b element A satisfy aRb, then [a] = [b], that is, the equivalence classes in which a and b lie are identical.") [a] = [b]

how can i now show that this is equivalent to $apb$?

furthermore how do i do question b?

Answer 1

HINT

You have to :

i) use the definition of Equivalence class : for a relation $R$ that is reflexive, symmetric and transitive :

$[a] = \{ x \in X \mid aRx \}$.

ii) to prove the equivalence : $[ a ] = [ b ]$ iff $aRb$, split the proof in two :

1) if $[ a ] = [ b ]$, then $aRb$

2) if $aRb$, then $[ a ] = [ b ]$.

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For 1) : if $[ a ] = [ b ]$, then - by definition - $b \in \{ x \in X \mid bRx \} = [ b ]$, because $bRb$ by reflexivity.

Thus, $b \in \{ x \in X \mid aRx \} = [ a ]$, i.e. $aRb$.