# Greatest lower bounds and least upper bounds

• Mar 31st 2008, 04:24 PM
mthrcks300
Greatest lower bounds and least upper bounds
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• Mar 31st 2008, 04:50 PM
ThePerfectHacker
Quote:

Originally Posted by mthrcks300
Let A be a partially ordered set. Suppose A is a subset of B and B is a subset of C. Assuming that all the least upper bounds and greatest lower bounds exist, prove that: glb(B)≤glb(A).

I start you off. If $\displaystyle x$ is the greatest lower bound for $\displaystyle B$ then $\displaystyle x\leq b$ for all $\displaystyle b\in B$. But then $\displaystyle x\leq a$ for all $\displaystyle a\in A$ because $\displaystyle A\subseteq B$. This means that $\displaystyle x\leq y$ where $\displaystyle y$ is greatest lower bound for $\displaystyle A$ because $\displaystyle x$ is a lower bound for $\displaystyle A$ and $\displaystyle y$ is the greatest lower bound for $\displaystyle A$.
• Mar 31st 2008, 05:27 PM
mthrcks300
You start by saying that x is the greatest lower bound for B, but then you say that y is the greatest lower bound for B. I'm confused.
• Mar 31st 2008, 06:31 PM
ThePerfectHacker
I made a mistake. Does it make more sense now?