## 1.3 Exercise 15

Yes, after months, I'm back ;-). This was an interesting problem to me because it explores a bit more deeply the concept of the LUBP.

"Assume that the real line has the least upper bound property.

(a) Show that the sets and have the least upper bound property.

(b) Does in the dictionary order have the least upper bound property? What about ? What about ?"

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

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SOLUTION

(a)

Pick any nonempty , bounded above in . Note that , and so , by transitivity of inclusion (a partial order axiom). Since has the LUBP (by assumption), the set of upper bounds of (in ), say , has a least upper bound, namely . Restricted to , the upper bounds are , and is still the LUB in . It follows that all nonempty subsets that are bounded above in have a least upper bound in , and has the LUBP. Following the exact same recipe works with all nonempty subsets of that are bounded above. As a point of clarification, notice that the set is not bounded above in , and so it is not an impediment to the base set having the LUBP.

(b)

Yes, in the dictionary order have the least upper bound property. Pick any nonempty subset of that is bounded above, as the interval , , , xor with of course. For all these, the set of upper bounds are , and the LUB is .

No, does not have the LUBP (in the dictionary order). We need only show a counterexample: take the interval (nonempty, bounded above) with and . The set of upper bounds is , and this clearly has no smallest element. In other words, by picking the smallest upper bound one could think of, say , the idea is that one can always come up with a (strictly) smaller one, say .

Yes, has the LUBP (in the dictionary order). Pick any nonempty subset of that is bounded above, as the interval , , , xor with and . For all these, is the LUB, since it's included in the upper bound set and it is its least element.