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Say we have a metric space $(M,d)$, and we want to complete it in the following sense:

Definition: A completion of $(M,d)$ is a complete metric space $(\widetilde{M},d')$ together with a Lipschitz funcion $i:M\rightarrow\widetilde{M}$ such that for every other complete metric space $(N,\rho)$ together with a Lipschitz function $f:M\rightarrow N$, there exists an unique Lipschitz function $F:\widetilde{M}\rightarrow N$ such that $F\circ i=f$.

This definition is adapted from the definition for uniform spaces (wikipedia). I changed the condition that the functions are uniformly continuous to Lipschitz so any two completions of a metric space would be "equivalent" as metric spaces, and not just be "uniformly equivalent". One could also suppose the functions are isometries, for example. (I think the better "morfisms" in the category of metric spaces are Lipschitz functions.)

The usual completion of $M$ is defined to be (a quotient of) the set $\widetilde{M}$ of Cauchy sequences on $M$ with the (pseudo)metric $d'\left((x_n)_n,(y_n)_n\right)=\lim d(x_n,y_n)$ and the inclusion $i:x\in M\mapsto (x)_n\in\widetilde{M}$ (it's the same as the one given in wikipedia).

However, suppose we are working with a normed (vector) space, let's say $(X,\Vert\cdot\Vert)$. The completion of $X$ (making the proper adaptations in the definition: that the functions are linear, etc...) can be very easily defined as the closure of $ev(X)$ as a subspace of $X''$, where $Y'$ denotes the dual of a normed space $Y$ with the operator norm, and $ev:X\rightarrow X''$ is the evaluation function: $ev(x)(f)=f(x)$ for every $x\in X$ and $f\in X'$.

My question is: would we be able to make a similar construction for general metric spaces, that is, to find a nice definition of dual of a metric space for which the dual of it would be a complete metric space? If so, we could try to just define the completion of $M$ as the closure of $ev(M)$ in $dual(dual(M))$. In other words, I would like to find a kind of (nice) (algebraic, analytic, etc..) structure so the category of sets with that structure is dual to the category of metric spaces.

The first possible "dual" of $(M,d)$ that comes to my mind is $C_b(M)$, the set of continuous bounded functions from $M$ to $\mathbb{R}$ with the $\infty$-norm. Problem is that this is a commutative, unital $C^*$-Algebra, and we know the natural dual of a commutative $C^*$-Algebra is a compact hausdorff topological space, not a metric space.

(also, I guess a dual notion of metric space couls have many applications other than just making completions)

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What is a "limited" function from $M$ to $\mathbb R$? I've never seen this term used before. Cool idea by the way. –  Patrick Da Silva Jul 3 '13 at 15:12
    
@PatrickDaSilva I'm almost sure it means "bounded". –  Daniel Fischer Jul 3 '13 at 15:51
    
@Daniel : Me too, but the word is still a bit weird. –  Patrick Da Silva Jul 3 '13 at 16:21
    
@Luiz : Another option for the dual would be the space of continuous functions with compact support ; I feel like it's a better option to understand what happens "around a point", it has less elements than $C_b(M)$ (if I assumed this is the space of continuous bounded functions). –  Patrick Da Silva Jul 3 '13 at 16:46
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Interesting question, but I am pretty sure that the answer is "No, there is no such construction." (unfortunately) –  Martin Brandenburg Jul 3 '13 at 21:41
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