# Relation of completeness to the l.u.b. property?

My book says that the real numbers are complete in the sense that they satisfy the least upper bound property. So it is the case that completeness and satisfying the l.u.b. property are equivalent by definition, or is it the case that satisfying the l.u.b. property implies completeness, meaning that there are other ways for an ordered field to be complete?

andrewkirk
Homework Helper
Gold Member
The word 'complete' has different definitions in different contexts. A metric space is complete if every Cauchy sequence in the space converges. In contrast, completeness for partially ordered sets is defined by reference to suprema or infima, of which the least upper bound property is an example.

Since the real numbers are both a metric space and a partially ordered set, the use of the term 'complete' is ambiguous, as each of the two possibilities gives a different definition of 'complete'. Fortunately, for the real numbers, the two definitions are logically equivalent. That's not the same thing as being equivalent by definition, as the definitions are different. But for the real numbers, satisfying one definition implies that the other is satisfied and vice versa.

More generally, for a metric space that is also a partially ordered set, I wonder if it can be proven that Cauchy completeness implies supremum completeness and/or vice versa.

Mr Davis 97
fresh_42
Mentor
I know completeness as "Every Cauchy sequences has a limit." defined on any ordered field, which means there are positive and negative numbers in this context. I've also found a proof that in ##\mathbb{R} \cup \{\pm \infty\}## both ##\displaystyle \mathop{\overline{\lim}}_{n \to \infty}## and ##\displaystyle \mathop{\underline{\lim}}_{n \to \infty}## are limit points, which I think generalizes to ordered fields in general. On the other hand, a metric on a topological space is unique up to isometries. Doesn't this already imply the equivalence of the two definitions for all metric spaces? Will say, I don't see what should have been to show.

Infrared
Gold Member
More generally, for a metric space that is also a partially ordered set, I wonder if it can be proven that Cauchy completeness implies supremum completeness and/or vice versa.

The order on the space might have nothing to do with the metric, so I don't see any hope for this being true in general. For example, take ##X=[0,1]## with the usual metric and use a bijection ##f:X\to (0,1)## to induce an order ##\leq_X## on ##X## defined by ##a\leq_Xb## if ##f(a)\leq f(b)##. Then ##X## is complete in the Cauchy sense but is isomorphic as an ordered set to ##(0,1)##, so not supremum complete. Conversely, you can consider ##(0,1)## as a metric space with an order defined by a bijection ##g:(0,1)\to [0,1]## as before to get a counterexample for the reverse implication.

On the other hand, a metric on a topological space is unique up to isometries.
I think this is false. Equipping ##\mathbb{R}## with ##d_1(x,y)=|x-y|## and ##d_2(x,y)=|\arctan(x-y)|## gives homeomorphic but non-isometric spaces. For a trivial example, take the two-element discrete topological space and give it metrics which assign different distances to the pair of distinct points.

fresh_42
Mentor
I think this is false. Equipping ##\mathbb{R}## with ##d_1(x,y)=|x-y|## and ##d_2(x,y)=|\arctan(x-y)|## gives homeomorphic but non-isometric spaces. For a trivial example, take the two-element discrete topological space and give it metrics which assign different distances to the pair of distinct points.
Correct, I confused the order. A completion to a given metric is unique (sort of) and not the other way around. Thanks for correction.

jim mcnamara
mathwonk