Let A be a set with a partial order R. For each a E A, let Sa={xEa:xRA} the section of a. Let F={Sa:aEA}. Then F is a subset of P(a) and thus may be partially ordered by (subset symbol).

Prove: if Sx is the least upper bound for {Sb:bEB} then B is subset of A and x is the least upper bound for B.

So this is how i started: Suppose Sx is the least upper bound. Then Sx is an upper bound for B and xRb for every upper bound b for B. Hence xEb Thus B is subset of A since xEa. This is where i got stuck

Describe explicitly the sets Sa, aEA, when A = R (the set of real numbers) and the partial order R is (less or equal symbol) " In this case show that set inclusion on F is a linear order. Do the same when A = N (the set of natural numbers), and the partial order R is "divides"." In this case show that set inclusion on F is not a linear order.

i have no clue how to this

Any help would be great

thanks