Greatest lower bounds and least upper bounds

**mthrcks300** · March 31st 2008, 04:24 PM

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**ThePerfectHacker** · March 31st 2008, 04:50 PM

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 $x$ is the greatest lower bound for $B$ then $x\leq b$ for all $b\in B$ . But then $x\leq a$ for all $a\in A$ because $A\subseteq B$ . This means that $x\leq y$ where $y$ is greatest lower bound for $A$ because $x$ is a lower bound for $A$ and $y$ is the greatest lower bound for $A$ .

**mthrcks300** · March 31st 2008, 05:27 PM

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.

**ThePerfectHacker** · March 31st 2008, 06:31 PM

I made a mistake. Does it make more sense now?

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