For problems 1 and 2 below, use the enclosed definitions of R1 and R2: Note: Thi

Question

For problems 1 and 2 below, use the enclosed definitions of R1 and R2:

Note: Think carefully about what make a pair of functions part of R1 (and similarly for R2) and what the difference is between R1 and R2.

1) Is R1 a partial order? Prove your answer.

2) Is R2 a partial order? Prove your answer.

Let (B, le B) be a poset and A be a set. The set FUN(A, B) is defined as the set of all functions with domain A and codomain B. Let R1 and R2 be relations on FUN(A, B) defined as follows: (f, g) epsilon R1 if and only if times epsilon A [f(x) le B g(x)] (f, g) epsilon R2 if and only if times epsilon A [f(x) le B g(x)]

Explanation / Answer

Note: Two functions f,g are equal if their domain(say A) is same and for all x in A f(x)=g(x)..

Proof:

we need to prove R1 is reflexive, anti symmetric and transitive

Reflexive: let f belongs to FUN(A,B) then for every x in A f(x)<=f(x) because B is a poset and f(x) belongs to B.So (f,f) belongs to R1 for every f in FUN(A,B).

Anti symmetric: let f,g belong to FUN(A,B), and suppose (f,g) and (g,f) belong to R1, then for all x in A f(x)<=g(x) and g(x)<=f(x) which means f(x)=g(x) as B is a poset.. So f=g by definition of equal functions.

Transitive: let f,g,h belong to FUN(A,B) and (f,g) in R1 and (g,h) in R1 then for all x f(x)<=g(x) and g(x)<=h(x) which means f(x)<=h(x) as B is a poset. So (f,h) belongs to R1.So R1 is transitive.

R2 is not partial order because anti symmetrric property fails...

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