## 1.3 Exercise 14

Rather than merely repeating the proof of 1.3.13 to show the converse, we can impose the "opposite" or symmetric order relation on the set to show that if it has the GLBP then it has the LUBP too.

"If is a relation on a set , define a new relation on by letting if .

(a) Show that is symmetric if and only if .

(b) Show that if is an order relation, is also an order relation.

(c) Prove the converse of the theorem in Exercise 13."

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

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SOLUTION

(a)

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Pick any . Suppose : this implies . Since , all elements of are also in . Thus , and is symmetric.

.

Pick so that . Also by symmetry of . Now, , and also , and is symmetric. Thus, we have the implication that via its symmetric partner, and : because all elements of are in , and is empty because we're carbon-copying the elements of into and was otherwise empty.

(b)

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Comparability. For , either or or . Thus, inherits comparability.

Nonreflexivity. For all , . Since such is not in , it could not have been generated in . Thus, , and inherits nonreflexivity.

Transitivity. For such that , where the presence of the first two implies the presence of the third. Next, we follow the implication for generating elements in set : . Rearranging, , and is transitive because the presence of the first two implied the presence of the third (via ).

(c)

First of all, if part (b) holds and , then is not symmetric (since if both , then by transitivity, contradicting nonreflexivity). Then, not symmetric implies by contrapositive of part (a). We've shown that we're talking about two different order relations.

Say has the LUBP GLBP, ordered as . This means that for *any nonempty* bounded above, the set of upper bounds, , has a least element, say , so that and . In particular, this means xor for all and xor for all . Additionally, since the LUBP implies the GLBP, for any nonempty set bounded below, the set of lower bounds, , has a greatest element, , and . This means xor for all and xor for all .

Now impose on , instead of , the order relation which is different. Following the implication for generating elements in this new order relation, xor , for all and xor for all . Thus, is a greatest lower bound. Also, xor for all , and xor for all . Thus is a least upper bound. Hence, if the set has the GLBP, it has the LUBP, as we wanted to show.