## Complete Uniform Spaces

Okay, in a uniform space we have these things called “Cauchy nets”, which are ones where the points of the net are getting closer and closer to each other. If our space is sequential — usually a result of assuming it to be first- or second-countable — then we can forget the more complicated nets and just consider Cauchy sequences. In fact, let’s talk as if we’re looking at a sequence to build up an intuition here.

Okay, so a sequence is Cauchy if no matter what entourage we pick to give a scale of closeness, there’s some point along our sequence where all of the remaining points are at least that close to each other. If we pick a smaller entourage we might have to walk further out the sequence, but eventually every point will be at least that close to all the points beyond it. So clearly they’re all getting pressed together towards a limit, right?

Unfortunately, no. And we have an example at hand of where it can go horribly, horribly wrong. The rational numbers are an ordered topological group, and so they have a uniform structure. We can give a base for this topology consisting of all the rays , the rays , and the intervals , which is clearly countable and thus makes second-countable, and thus sequential.

Okay, I’ll take part of that back. This is only “clear” if you know a few things about cardinalities which I’d thought I’d mentioned but it turns out I haven’t. It was also pointed out that I never said how to generate an equivalence relation from a simpler relation in a comment earlier. I’ll wrap up those loose ends shortly, probably tomorrow.

Back to the business at hand: we can now just consider Cauchy sequences, instead of more general Cauchy nets. Also we can explicitly give entourages that comprise a base for the uniform structure, which is all we really need to check the Cauchy condition: . I did do absolute values, didn’t I? So a sequence is Cauchy if for every rational number there is an index so that for all and we have .

We also have a neighborhood base for each rational number given by the basic entourages. For each rational number we have the neighborhood . These are all we need to check convergence. That is, a sequence of rational numbers converges to if for all rational there is an index so that for all we have .

And finally: for each natural number there are only finitely many square numbers less than . We’ll let be the largest such number, and consider the rational number . We can show that this sequence is Cauchy, but it cannot converge to any rational number. In fact, if we had such a thing this sequence would be trying to converge to the square root of two.

The uniform space is shot through with holes like this, making tons of examples of Cauchy sequences which “should” converge, but don’t. And this is all just in one little uniform space! Clearly Cauchy nets don’t converge in general. But we dearly want them to. If we have a uniform space in which every Cauchy sequence *does* converge, we call it “complete”.

Categorically, a complete uniform space is sort of alike an abelian group. The additional assumption is an extra property which we may forget when convenient. That is, we have a category of uniform spaces and a full subcategory of complete uniform spaces. The inclusion functor of the subcategory is our forgetful functor, and we’d like an adjoint to this functor which assigns to each uniform space its “completion” . This will contain as a dense subspace — the closure in is the whole of — and will satisfy the universal property that if is any other complete uniform space and is a uniformly continuous map, then there is a unique uniformly continuous extending .

To construct such a completion, we’ll throw in the additional assumption that is second-countable so that we only have to consider Cauchy sequences. This isn’t strictly necessary, but it’s convenient and gets the major ideas across. I’ll leave you to extend the construction to more general uniform spaces if you’re interested.

What we want to do is identify Cauchy sequences in — those which should converge to something in the completion — with their limit points in the completion. But more than one sequence might be trying to converge to the same point, so we can’t just take *all* Cauchy sequences as points. So how do we pick out which Cauchy sequences should correspond to the same point? We’ll get at this by defining what the uniform structure (and thus the topology) *should* be, and then see which points have the same neighborhoods.

Given an entourage of we can define an entourage as the set of those pairs of sequences where there exists some so that for all and we have . That is, the sequences which get eventually -close to each other are considered -close.

Now two sequences will be equivalent if they are -close for all entourages of . We can identify these sequences and define the points of to be these equivalence classes of Cauchy sequences. The entourages descend to define entourages on , thus defining it as a uniform space. It contains as a uniform subspace if we identify with (the equivalence class of) the constant sequence . It’s straightforward to show that this inclusion map is uniformly continuous. We can also verify that the second-countability of lifts up to .

Now it also turns out that is complete. Let’s consider a sequence of Cauchy sequences . This will be Cauchy if for all entourages there is an so that if and the pair is in . That is, there is an so that for and we have . We can’t take the limits in of the individual Cauchy sequences — the limits along — but we *can* take the limits along ! This will give us another Cauchy sequence, which will then give a limit point in .

As for the universal property, consider a uniformly continuous map to a complete uniform space . Then every point in comes from a Cauchy sequence in . Being uniformly continuous, will send this to a Cauchy sequence in , which must then converge to some limit since is complete. On the other hand, if is another representative of then the uniform continuity of will force , so is well-defined. It is unique because there can be only one continuous function on which agrees with on the dense subspace .

So what happens when we apply this construction to the rational numbers in an attempt to patch up all those holes and make all the Cauchy sequences converge? At long last we have the real numbers ! Or, at least, we have the underlying complete uniform space. What we don’t have is any of the field properties we’ll want for the real numbers, but we’re getting close to what every freshman in calculus thinks they understand.

[…] Miscellany Well, yesterday was given over to exam-writing, so today I’ll pick up a few scraps I mentioned in passing on Thursday. […]

Pingback by Miscellany « The Unapologetic Mathematician | December 1, 2007 |

[…] Order on the Real Numbers We’ve defined the real numbers as a topological field by completing the rational numbers as a uniform space, and then extending the field operations to the new points […]

Pingback by The Order on the Real Numbers « The Unapologetic Mathematician | December 5, 2007 |

[…] functions are defined they agree. Since the rationals are dense in the reals (the latter being the uniform completion of the former) there can be only one continuous extension of to the whole real line. We’ll […]

Pingback by Exponentials and Powers « The Unapologetic Mathematician | April 11, 2008 |

[…] We defined the real numbers to be a complete uniform space, meaning that limits of sequences are convergent if and only if they are Cauchy. Let’s write […]

Pingback by Cauchy’s Condition « The Unapologetic Mathematician | April 23, 2008 |

[…] Now we’ve gone through a lot of work to just add one little extra element to our field, but it turns out this is all we need. Luckily enough, the complex numbers are already algebraically complete! This is very much not the case if we were to try to algebraically complete other fields (like the rational numbers). Unfortunately, the proof really is essentially analytic. It seems to be a completely algebraic statement, but remember all the messy analysis and topology that went into defining the real numbers. […]

Pingback by The Complex Numbers « The Unapologetic Mathematician | August 7, 2008 |

[…] we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the […]

Pingback by Power Series « The Unapologetic Mathematician | August 18, 2008 |

[…] integral does not exist. Yes, is countable, and so it has measure zero. However, it’s also dense, which means is discontinuous everywhere in the unit […]

Pingback by Iterated Integrals III « The Unapologetic Mathematician | December 18, 2009 |

[…] a lemma: if is irrational, then the set of all numbers of the form with and any integers is dense in the real line. That is, every open interval contains at least one point of . The same is true […]

Pingback by Non-Lebesgue Measurable Sets « The Unapologetic Mathematician | April 24, 2010 |

[…] got a normed vector space, it’s a natural question to ask whether or not this vector space is complete or not. That is, we have all the pieces in place to define Cauchy sequences in our vector space, […]

Pingback by Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces « The Unapologetic Mathematician | May 12, 2010 |

[…] the real numbers form a complete metric space, being Cauchy and being convergent are equivalent — a sequence of finite real numbers is […]

Pingback by Convergence Almost Everywhere « The Unapologetic Mathematician | May 14, 2010 |

[…] We won’t talk quite yet about convergence because our situation is sort of like the one with rational numbers; we have a sense of when functions are getting close to each other, but most of these mean Cauchy […]

Pingback by The L¹ Norm « The Unapologetic Mathematician | May 28, 2010 |

[…] We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring is […]

Pingback by Completeness of Boolean Rings « The Unapologetic Mathematician | August 10, 2010 |

[…] old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space is a Baire space. Let be a countable collection of open dense subsets of . We will show that […]

Pingback by Baire Sets « The Unapologetic Mathematician | August 13, 2010 |

[…] every nonempty set of numbers with an upper bound has a least upper bound. We’ve also used Cauchy sequences to “complete” the uniform structure on the rational numbers — to make sure that […]

Pingback by Cuts and Sequences are Equivalent « The Unapologetic Mathematician | January 10, 2015 |