Complete metric space

Metric completion

The completion of a metric space may be thought of as the smallest possible metric space containing but with all limits added, #m/def/anal i.e. a complete metric space. This notion is made rigorous by the universal property, which ensures uniqueness up to unique isomorphism.

Universal property

The metric completion is characterized up to unique isomorphism in by the following universal property:

is complete. If is a complete metric space and such that is dense in , then there exists a unique isometry such that , i.e. the following diagram commutes

https://q.uiver.app/#q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiXFxiYXIgWCJdLFsyLDIsIlkiXSxbMCwxLCJcXGlvdGEiXSxbMSwyLCJcXGJhciBmIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzAsMiwiZiIsMl1d

This of course forms a Free-forgetful adjunction into the full subcategory of complete metric spaces and isometries.

Construction

Let denote the set of all Cauchy sequences on , For any sequences , let

which defines an equivalence relation. The completion is the quotient with a metric given by

and .

Validity of construction

First we will show that indeed defines an equivalence relation. reflexivity and symmetry are clear, and transitivity follows from triangle inequality: If and , then so .

Next we show that the metric on is well-defined. Let and . Then by the triangle inequality

but by symmetry the reverse inequality holds too, so . symmetry and positive definite follow immediately. For positive definite note

as required.

Now we need to show that the given construction is indeed complete. We make the following observations

  • Let . If , then so too is every subsequence .
  • Let . Since is Cauchy, for every there exists a subsequence such that for all .

Let be a Cauchy sequence in . For each , we fix a representative in such that for all we have . Since is Cauchy, for every there exists an such that for all we have (using that sufficiently large have ). Now let which is Cauchy in , since for any if and thus we have

We claim is the limit of .

Let and . Then

so .

It remains to show that satisfies the universal property. Let be a complete metric space and .

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