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  [0,1] = \{x \ \vert \ 0 \leq x \leq 1\} and  [0,1) = \{x \ \vert \ 0 \leq x < 1\} have the least upper bound property.

(b) Does  [0, 1] \times [0, 1] in the dictionary order have the least upper bound property?  What about  [0,1] \times [0, 1) ?  What about  [0,1) \times [0,1] ?"

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

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SOLUTION

(a)

Pick any nonempty  A_0 \subset [0,1] , bounded above in  [0,1] .  Note that  A_0 \subset [0,1] \subset \mathbb{R} , and so  A_0 \subset \mathbb{R} , by transitivity of inclusion (a partial order axiom).  Since  \mathbb{R} has the LUBP (by assumption), the set of upper bounds of  A_0 (in  \mathbb{R} ), say  [b, \infty) , has a least upper bound, namely  b .   Restricted to  A = [0,1] , the upper bounds are  [b, \infty) \cap [0, 1] = [b, 1] , and  b is still the LUB in  A .  It follows that all nonempty subsets that are bounded above in  A have a least upper bound in  A , and  A has the LUBP.  Following the exact same recipe works with all nonempty subsets of  A' = [0,1) that are bounded above.  As a point of clarification, notice that the set  A_0' = A' = [0, 1) is not bounded above in  A' , and so it is not an impediment to the base set  A' having the LUBP.

(b)

Yes,   [0, 1] \times [0, 1] in the dictionary order have the least upper bound property.  Pick any nonempty subset of  A = [0, 1] \times [0, 1] that is bounded above, as the interval  (x_1 \times y_1, x_2 \times y_2) ,  [x_1 \times y_1, x_2 \times y_2) ,  (x_1 \times y_1, x_2 \times y_2] , xor  [x_1 \times y_1, x_2 \times y_2] with  x, y \in [0,1] of course.  For all these, the set of upper bounds are  [x_2 \times y_2, 1 \times 1] , and the LUB is  x_2 \times y_2 .

No,  [0,1] \times [0, 1) does not have the LUBP (in the dictionary order).  We need only show a counterexample: take the interval   (x_1\times y_1, x_1 \times 1) (nonempty, bounded above) with  x \in [0, 1] and  y \in [0, 1) .  The set of upper bounds is  (x_1 \times 0, 1 \times 1) , and this clearly has no smallest element.  In other words, by picking the smallest upper bound one could think of, say  x_2 \times 0 , the idea is that one can always come up with a (strictly) smaller one, say  \frac{x_2}{2} \times 0 .

Yes,  [0,1) \times [0,1] has the LUBP (in the dictionary order).  Pick any nonempty subset of  A = [0, 1) \times [0, 1] that is bounded above, as the interval  (x_1 \times y_1, x_2 \times y_2) ,  [x_1 \times y_1, x_2 \times y_2) ,  (x_1 \times y_1, x_2 \times y_2] , xor  [x_1 \times y_1, x_2 \times y_2] with  x \in [0, 1) and  y \in [0, 1] .  For all these,  x_2 \times y_2 is the LUB, since it's included in the upper bound set and it is its least element.

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