# [Set Theory] Proving Linear Order on a Subset of a Partially Ordered Set

EstimatedEyes

## Homework Statement

Let L be a partially ordered set. Every countable chain in L has an upper bound. Let S be a countable subset of L such that for arbitrary a,b in S there exists a c in S such that a (less-than-or-equal) c and b (less-than-or-equal) c. Prove S has an upper bound in L.

## Homework Equations

Definition of linear, partial orders.

## The Attempt at a Solution

I assume the goal is to prove that S is linearly ordered and thus a chain, having an upper bound in L by the problem text. S is clearly a partial order because it is a subset of L. After listing the properties of a partial order, I have no idea how to continue. Any help would be greatly appreciated! Thanks!

Staff Emeritus
Homework Helper
S is not a chain in general. So you won't be able to prove that. But try selecting a few elements in S which DO form a chain. This chain will have an upper bound. And if you selected the elements correctly, then the upper bound of the chain, will be an upper bound of S...

EstimatedEyes
Okay, I assume that a,b, and c are the elements to which you are referring. Obviously a and b are comparable with c but I am having trouble showing that it then follows that a is comparable with b. I know it can't be this easy, but I don't completely understand why c is not already an upper bound because it is defined to be greater than or equal to all elements of S. Since S is a subset of L, c must also be an element of L, giving S an upper bound in L. What is wrong with this line of thinking?

Staff Emeritus