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

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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?
 

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  • #2
andrewkirk
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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.
 
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  • #3
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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.
 
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Infrared
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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.
 
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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.
 
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  • #6
mathwonk
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This was a confusing point to me, as I had heard there is only one complete ordered field, namely the real numbers, but then I saw examples of other "complete" non archimedean ordered fields. A correct statement seems to be that there is only one complete archimedean ordered field. The confusion occurs because of the two meanings of "complete". The lub version of complete also implies archimedean, whereas the Cauchy version of complete does not. So indeed there is only one lub-complete ordered field. Google some examples of complete non archimedean ordered fields. They tend to involve Laurent series. To get an example start from any non archimedean ordered field and form the Cauchy completion. This is discussed in standard old algebra books from my student days like Van der Waerden and Lang, but perhaps not in more modern ones like Dummitt and Foote.
 
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