 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/calculus-probability-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/7/c/solution

Take the natural numbers including zero with their usual total order:

P = ({0, 1, 2, \dots}, \leq) .

A nonempty subset with an upper bound

m \in P

is contained in the finite set

{0, 1, \dots, m}

0

. Thus

P

But

D = P

max (x, y) \in P

is above any two members. It has no upper bound in

P

: for any proposed bound

m

, the member

m + 1

is larger. It therefore has no least upper bound. Hence

P is bounded-complete but not directed-complete .

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