Let a be a set with a partial order r. for each a∈a, let sa= {x∈a: xra}. let f={sa: a∈a}. then f is a subset of p(a) and thus may be partially ordered by ⊆, inclusion.

a) show that if arb, then sa ⊆ sb.
b) show that if sa ⊆ sa, then arb.
c) show that if b⊆a, and x is the least upper bound for b, then sx is the least upper bound for {sb: b∈b}

See proofs below

Step-by-step explanation:

a) Suppose that aRb. Let y∈Sa , then y∈A and yRa. We have that yRa and aRb. Since R is a partial order, R is a transitive relation, therefore yRa and aRb imply that yRb. Now, y∈A and yRb, thus y∈Sb. This reasoning applies for all y∈Sa that is, for all y∈Sa, y∈Sb, threrefore Sa⊆Sb.

b) Suppose that Sa⊆Sb. Since R is a partial order, R is a reflexive relation then aRa. Thus, a∈Sa. The inclusion Sa⊆Sb implies that a∈Sb, then aRb.

c) Denote this set by S={Sb:b∈B}. We will prove that supS=Sx with x=supB.

First, Sx is an upper bound of S: let Sb∈S. Then b∈B and since x=supB, x is an upper bound of B, then bRx. Then b∈Sx. Now, for all y∈Sb, yRb and bRx, then by transitivity yRx, thus y∈Sx. Therefore Sb⊆Sx for all Sb∈S, which means that Sx is an upper bound of S (remember that the order between sets is inclusion).

Now, let's prove that Sx is the least upper bound of S. Let Sc⊆A be another upper bound of S (in the set F). We will prove that Sx⊆T.

Because Sc is an upper bound, Sb⊆Sc for all b∈B. Thus, if y∈Sb for some b, then y∈Sc. That is, if yRb then yRc. In particular, bRb then bRc for all b∈B. Thus c is a upper bound of B. Byt x=supB, then xRc. Now, for all z∈Sx, zRx and xRc, which again, by transitivity, implies that z∈Sc. Therefore Sx⊆Sc and Sx=sup S.

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step-by-step explanation:

answer: -8.2562234e+13