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Archive for March 2nd, 2009

## 1.3 Exercise 13

This theorem (and its converse) is needed everywhere in analysis and topology, and it is very important.

"Prove the following:

Theorem.  If an ordered set $A$ has the least upper bound property, then it has the greatest lower bound property."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)

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SOLUTION

Suppose that $A$ does not have the GLBP. This means that there exists a nonempty $A_0 \in A$ that is bounded below but does not have a greatest lower bound, so that the set of lower bounds $B_l \in A$ (nonempty) does not have a largest element.  Additionally, $A_0$ does not have a smallest element, for, if it did, this would in fact be its infimum (such an element would be in the set of lower bounds because it is lesser or equal to all elements of $A_0$).  Next let's focus on $B_l$, which is a set in $A$ as well.  It cannot possibly have the LUBP, because it has no greatest element and because $A_0$, which now bounds it above, has no smallest element. We found a nonempty subset of $A$ with no supremum, and we've proved the theorem by contrapositive.

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