In November of last year, I published a guide, if you will, on how to take an arbitrary metric space and obtain a complete metric space which contains a dense isometric “copy” of . This process is called *completing a metric space*, and the space is called the *completion* of .

The construction of that I gave involved defining an equivalence relation between Cauchy sequences in by

We then defined to be the set of equivalence classes endowed with the metric

,

where and are representative elements of the equivalence classes and , respectively. has a natural embedding into by sending each to the equivalence class containing the sequence which is identically .

Today, I want to give an entirely different construction using Lipschitz functions. This construction came to my attention by way of Halsey Royden’s textbook *Real Analysis. *The construction is actually part of Exercise 17 in Section 13.2.

Let be a nonempty metric space containing a point . Denote the space of Lipschitz functions which vanish at the point by . I claim that is a real normed space, with norm defined by

Indeed, fix real scalars and functions . It is clear that the function vanishes at the point . Since and are Lipschitz,there exist positive constants and such that

Set . Then

To see that is a norm, observe

Since we require that , we conclude that is the zero function. Positive homogeneity is obvious, and the triangle inequality is a consequence of the triangle inequality of absolute value.

Moreover, is a Banach space. Indeed, let be a Cauchy sequence in . I claim that, for any point , the real sequence is Cauchy. Indeed, let and be sufficiently large so that, for given,

,

where we use the hypothesis that . Hence,

So, the sequence has a pointwise limit . It remains for us to show that . Since for all , it is clear that . To see that is Lipschitz on the space , observe that

Above, we used the result that converges in , since the reverse triangle inequality implies it is a Cauchy sequence. We will later see that . Lastly, let be a natural number such that implies that . For any with ,

Since this bound is independent of and , we can take the supremum of the LHS above over all to obtain .

I claim that there is an isometric embedding of into the space of bounded linear functionals on (i.e. the dual space of ), which we denote by . Indeed, for each , let denote the linear functional which evaluates the function at the point . I claim that

Fix . Then by definition of the norm,

For the reverse inequality, observe that the map

is a Lipschitz function which vanishes at . Since if and only if , we see that the mapping is injective. Since any closed subset of a complete metric space is also complete, we conclude that the closure of is a completion of .

We actually didn’t need to show that is a Banach space in order to prove that has a completion in , since the space of bounded (real or complex) linear functionals is Banach space, even if is not complete.

# References

Halsey Royden and Patrick Fitzpatrick, *Real Analysis **(Fourth Edition), *Prentice Hall, New York, 2010.

Reblogged this on Being simple and commented:

very interesting post about completation of a metric space.

Thanks for the reblog and the kind words!

actually misprint in completion