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Let $R$ be a reflexive and transitive relation on a set $A$. Define a new relation $E$ on $A$ as

$E=\{(a, b) \mid (a, b) \in R \text{ and } (b, a) \in R \}$

Prove that $E$ is an equivalence relation on $A$.

Define a relation $\leq$ on the equivalence classes of $E$ as $E_1 \leq E_2$ if $\exists a, b$ such that $a \in E_1, b \in E_2 \text{ and } (a, b) \in R$. Prove that $\leq$ is a partial order.
asked in Set Theory & Algebra by Veteran (59.5k points)
edited by | 490 views

2 Answers

+5 votes
Best answer

1. Since it is given that relation $R$ is reflexive and transitive, the new defined relation (definition of symmetric) is equivalence 

2. Partial order is a binary relation "≤" over a set P which is reflexiveantisymmetric, and transitive

answered by Boss (10.9k points)
edited by
0

@Rajesh Pradhan

@asu

but the definition of symmetric is :-

${ (a,b) \in R \rightarrow (b,a) \in R}$

Moreover if A = {1,2}

Relation R = {(1,2)} is symmetric, but according to above definition it is not.

Please clear this doubt.

+2

reflexive and transitive does nt mean symmetric ... 

0
please explain (b) part with example
0
What are the equivalence classes of E here?
+2 votes

for part 1

Since it is given that relation R is reflexive and transitive but we can't say anything about symmetricity of R.In the worst case, R is not symmetric then relation E contains diagonal elements which leads to an equivalence relation.

answered by Loyal (6.7k points)
edited by
+1

Ans b. E1≤E2 if ∃a,b such that a∈E1,b∈E2 and (a,b)∈R says there must be some pair (a,b) such that it is not existing as a SYMMETRIC PAIR in R i.e. (b,a) not ∈R. Only then it is possible that (a,b) ∈R goes into two different equivalence classes of E. 

Simply,  if ∃a,b such that a∈E1,b∈E2 and (a,b)∈R THEN (b,a) not ∈R

0
please give example for (b) part


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