- #1

pandaBee

- 23

- 0

## Homework Statement

Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least

upper bound of B1, and x2 is the least upper bound of B2. Prove that if

B1 ⊆ B2 then x1Rx2.

## Homework Equations

## The Attempt at a Solution

I split the proof into two different cases:

case 1: x_1 is an element of B_1, therefore (x_1 R x_2)

case 2: x_1 is not an element of B_1 then there are two subcases:

case 2a: x_1 is an element of B_2, therefore (x_1 R x_2)

case 2b: x_1 is not an element of B_2, therefore since x_1 is the least upper bound of B_1 and since x_1 is in neither B_1 or B_2, x_1 is an upper bound of B_2, in fact x_1 is the least upper bound of B_2 therefore (x_1 R x_2) by reflexivity of partial orders.

It's pretty straightforward for cases 1, and 2a, but for 2b how would I represent this mathematically? In my proof it relies on the fact that x_1 is the lowest upper bound of B_2 when

a) it is a lowest upper bound of B_1

b) it is not in either B_1 or B_2

c) B_1 is a subset of B_2

I can understand why it works, but I am just unsure of how to write it down.