Here’s a summary from the last post regarding completeness:

**Completeness: **A complete space is one in which all sequences that *want *to converge *do *converge. Intuitively, this means that if “points” (I use this term loosely, because points could represent continuous functions, or categories, etc) in a sequence eventually become arbitrarily close with respect to some sort of metric, then there will, in fact, be a limit point contained in the space to which the sequence converges.

With that in mind, I’ll proceed to give a fairly comprehensive discussion of completeness.

**(Pretty Much) Everything You’d Want to Know About Completeness**

Put formally, a complete space is one in which all Cauchy sequences converge. Named after the French mathematician Augustin-Louis Cauchy, a Cauchy sequence formalizes what is means for points to “get really close”.

Definition 1:Let be a metric space. We call a sequenceCauchyif, for any , there exists so that, for , we have .

So what’s the difference between a Cauchy sequence and a convergent sequence? Well, a Cauchy sequence need not always converge, whilst a convergent sequence is always Cauchy. To see this, consider a convergent sequence . By the definition of convergence, there exists so that for , we have that and . By the triangle inequality, this implies that for all . Thus, this sequence is Cauchy.

One of the most important aspects of complete spaces is the fact that we can prove convergence of a sequence without actually knowing *to what *that sequence converges. For example, if we have a sequence of functions , I can prove that this sequence converges (pointwise) simply by showing that . Thus, since is complete, this sequence will converge. This is remarkably useful in, say, the theory of normed spaces, since and therefore a Cauchy sequence of norms will converge.

The reader should note, **emphatically, **that completeness is not a topological property. That is, two topological spaces may be homeomorphic, yet one space may be complete, while the other is not. For example, let be the Euclidean metric and let be the discrete metric. Define . Clearly is not complete (since it doesn’t contain 0), yet there exists a homeomorphism and one may check that is complete. In fact, any metric space equipped with the discrete metric is complete. To see this, take any Cauchy sequence in a metric space equipped with the discrete metric. Then, for every , there is an so that but, by definition of the discrete metric, if , then so that the sequence eventually becomes constant after finitely many terms, and thus a constant sequence is trivially convergent.

Suppose I take a subset of a complete metric space. How are we to know if this subset, taken as its own metric space, is also complete? Well the following theorem characterizes precisely this.

Theorem 1:Let be a complete metric space. Let . is complete if and only if is closed in .

Proof:Suppose is closed. Then, given a Cauchy sequence in , this sequence has a limit because is complete. Since is closed, it contains its limit points so that and therefore is complete.Conversely, suppose is a complete subset of . Then, we need show that contains its limit points. Take . By definition of a limit point, there is a sequence converging to with elements in . Since a convergent sequence is a Cauchy sequence, this sequence has limit in by completeness. By uniqueness of limits (since, in fact, metric spaces are Hausdorff), this limit in must be . Ergo, implies so that is closed. QED.

As a rather simple example, consider as a complete metric space. Then, say, the open set is not complete as a metric space, yet the closed set is a complete metric space. This also begs the question: when does changing the metric alter the completeness of a space? Recall the definition of equivalent metrics.

Theorem 2:Let be a complete metric space. If , then is a complete metric space.

Proof idea:Since two equivalent metrics differ by a constant, if one space is complete, then the other is complete, just “closer in” or “further out” in each Cauchy sequence. This will become clear formally.

Proof:Since , there exists so that, for any we have . Thus, a Cauchy sequence with respect to , that is, is the same as . Since is complete, such a sequence converges and therefore converges in . QED.

For those readers with knowledge of Banach Spaces (I’ll write about these eventually) this means that equivalent norms preserve the structure of a Banach space.

Another interesting property of complete spaces is the fact that they *inherit* completeness from a dense subset.

Theorem 3:Let be a metric space. Let be dense subset of such that every Cauchy Sequence in converges in . Then, is complete.

Proof idea:We’re going to construct a sequence of points in that get really close to a sequence points in , and show that this is indeed a Cauchy sequence in and therefore converges. Finally, we show than the sequence in converges to the same limit.

Proof:Let be a Cauchy sequence in . Thus, there exists such that, for , . Since is dense in , for each , we may find such that .

Let such that, for , we have . Let . Then, for , we see (by the triangle inequality and hypotheses) that

Thus, is Cauchy and, by supposition, converges to . We shall show that .

Let be such that, if , . Let be as above. Then,

Therefore, and so is complete. QED.

I’m going to use this theorem for one proof that every metric space has a *completion. *A completion is a way of “making a metric space complete” by identifying it with a dense subset of a complete metric space.

Definition 2:Acompletionof a metric space is a complete metric space along with an isometry (distance-preserving function) so that is dense in .

For example, the completion of is simply . What we wish to see is that *all *metric spaces do indeed have a unique completion up to isometry. The first proof I’ll give is the “slick” proof using the completeness of the space of bounded continuous functions. If the reader is unfamiliar with the notion of uniform convergence and the space , an alternate proof is provided. A discussion of the space of bounded continuous functions can be found here.

Theorem 4:Every metric space has a completion.

Proof:Let be a metric space. Fix and define a mappingby

This function is continuous as it is the difference of two continuous functions (see Basic Continuity) and is bounded by the reverse triangle inequality . Therefore, .

Moreover, is an isometry, since, given , we have

.

Ergo, taking the completion of the image of we obtain a closed subset of , a complete metric space, and therefore a complete subset. QED.

This is the more elegant proof, but I’m also going to write out the much more involved “brute force” proof. So, let’s do that.

Theorem 4:Every metric space has a completion.

Proof:Let be a metric space.Call two Cauchy sequences and in equivalent if . I leave it to the reader to check that this indeed satisfies the axioms of an equivalence relation.

Let be the set of all equivalence classes so obtained. If , and , and , define

.

To see that defines a metric on , we need show that is unchanged if and are replaced by equivalent sequences.

Let . Given , let and . We have

Note thatso the right hand side is just equal to by the definition of the equivalence. Do the same for the other elements in and to see the two inequalities

and

Combining these inequalities, we get that so that is a well-defined metric on .

For each , there is a Cauchy sequence all of whose terms are . Let be the element of which contains this sequence. We need show that for all . In other words, the mapping is an isometry .

This is simple. Let . Then, Thus, is an isometry.

So, what we’re doing here is working towards applying Theorem 3. We have an isometry. Check. Now we need to show that is dense in , and that every Cauchy sequence in converges in .

We need show that, given any element , and any , there is an element such that . Choose . Since this is Cauchy, choose , such that, if , . Apply to so that . Then, Thus, is dense in .

We now apply Theorem 3 to .

Thus, we need find a Cauchy sequence in that converges in . Let be a Cauchy Sequence in . Then, given , there is such that, for , we have . For each , let . Then is Cauchy, since is an isometry. Let . Again, since is an isometry, . Thus, . Ergo, by the Theorem 3, is a complete metric space. So, we are (finally) done. QED.

So, we’ve given two proofs that all metric spaces do indeed have a completion. I’ll end with a statement and proof of Baire’s Theorem, a vitally important theorem in general topology and functional analysis.

Theorem 5:If is a complete metric space, then the intersection of a countable number of dense subsets is again a dense subset.

Proof idea:What we’re going to do is create a Cauchy sequence in the intersection, and use completeness of to show that this converges and so that the intersection is dense.

Proof:Let be a countable collection of open dense subsets. Let be an arbitrary open subset in . We need show that . Consider . This is an open set as the intersection of two open sets, so there exists and so that for the closed ball .

Now, assume that and have been chosen, then, is again open by density of , so we may choose so that . Thus, given any , choose . Then, for , we see that . So, by the triangle inequality,

.

We therefore conclude that is Cauchy, so it converges by the completeness of . You can check that this means for each . Finally, we conclude that or that the intersection is indeed dense, since was arbitrary. QED.

This theorem is used, for example, to prove the Open Mapping Theorem and the Closed Graph Theorem in functional analysis.

I think that’s about all I have to say about completeness for now; in the last of these three posts, I’ll cover connectedness in both metric spaces and topological spaces. If I think of anything else related to completeness that may be of interest, I will add it later. If you want to see anything added, as usual, contact erdosninth@gmail.com

Cheers,

J.T. (S.D.)