## 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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