# Partial order relation: lub and glb

• Dec 17th 2012, 11:20 PM
nicnicman
Partial order relation: lub and glb
I'm a little confused by this problem:

Let D = {2, 3, 6, 8, 9, 10, 12, 15, 18, 20} and R is the partial order relation defined on D where a R b means a | b.

Find the requested elements if they exist.

1. The lub({3,10})
2. The lub({2, 9})
3. The glb({12, 10})

1. lub({3,10}) does not exist.
2. The lub({2, 9}) = 18.
3. The glb({12, 10}) = 2.

Am I doing this correctly? Any suggestions are welcome.
• Dec 17th 2012, 11:26 PM
jakncoke
Re: Partial order relation: lub and glb
Yes. You are

Basically, partial order also can be symbolically represented as $\leq$ So when it means, find the lub for a subset $A \subset B$
Basically find the smallest number $b \in B$ such that for any $a \in A$ $a \leq b$ such the $\leq$ represents division, this means find the smallest element in B which is divided by everything in A and glb means find the greatest element in B which divides every element in A. But you seem to know all this already since you did the problems right.
• Dec 17th 2012, 11:35 PM
nicnicman
Re: Partial order relation: lub and glb
Nice. Thanks.