Completeness I

January 1, 2016

Last semester, I helped a friend review McGill's Analyis 3 course by trying to provide a better feel for completeness; this post will be a slightly edited version of that. Originally, I wanted to write about both completeness and compactness, and their connections, but I ended up only getting to completeness, and actually not everything I wanted to talk about. So I'll call this Completeness I, leaving open the possibility for more of these in the future. Hopefully the style is casual and expository without being misleading.

Completeness

In any metric space, every convergent sequence is Cauchy: if the points are clustering around the limit point, then they must become close.

On the other hand, if a sequence is Cauchy and has a convergent subsequence, then it is convergent: If all terms are close, and some terms are eventually close to the limiting value , then all terms are eventually close to the limiting value .. This is an equivalence of course: if a sequence is convergent, then it is both Cauchy and has a convergent subsequence.

A metric space is complete if every Cauchy sequence converges. We will see that completeness means what it's name suggests: there are no missing bits. We immediately have then have that a metric space is complete if every sequence has a convergent subsequence. This condition is called sequential compactness; we just saw that sequential compactness implies completeness, but completeness alone will not guarantee us sequential compactness.

What does it mean?

We have a condition, completeness; let us start by seeing how well-behaved it is. First, and are complete spaces iff is complete, since both Cauchyness and convergence just pass to the components. More interestingly, are subspaces of complete metric spaces also complete?

Proposition. Consider a complete metric space , and a subspace . Then is complete if and only if is a closed subset of .

Proof. The proof goes by noticing that the property of Cauchyness does not require a limiting value, and only depends on the elements in the sequence. So a sequence that has each element in is Cauchy in iff it is Cauchy in . The connection to closedness is that to be closed means to contain all of the limit points. So we start by considering as a subset of , and looking at its limit points and thus sequences which converge in but whose elements are contained in . Then since Cauchyness is equivalent to convergence in , we pass to Cauchy sequences, which then lets us consider sequences in . The proof is short, and feels a little magical to me.

This is our first hint that completeness means we aren't missing bits, and also what kind of "bits'' we might be missing: the only way to lose completeness by getting rid of some of our space is by lacking limit points.

We can look at a similar type of bit we might be missing: the infinite intersection of closed sets. We say a sequence of sets is contracting if

This leads us to a theorem:

Theorem (Cantor Intersection Theorem). Given a metric space , the following are equivalent:

  1. is complete.
  2. For any contracting sequence of non-empty closed subsets of , there exists such that

Proof sketch. To show , we simply take a sequence and show it's Cauchy, using the contracting condition. Completeness gives us a limit point, and since each set is closed, the limit point is in each . A final application of the contracting condition gives us uniqueness.

On the other hand, given a Cauchy sequence , we can define a contracting being Cauchy gives us the contracting condition. sequence of closed sets by ; then (2) gives us convergence.

There are more ways complete spaces are determined by missing bits. Let's upgrade to a normed space.

Proposition. Let be a normed space. Then the following are equivalent:

  1. is complete.
  2. If for are such that , then there exists with Recall that means .

Proof sketch. To show , we just note that the partial sums form a Cauchy sequence using the absolute convergence of the norms, and hence converge to some .

For , consider a Cauchy sequence in . By our earlier remarks, we only need to find a convergent subsequence. Choose For a general Cauchy sequence, we don't know how fast the terms get close together. So we exploit that they do get close to choose a subsequence where the terms get close exponentially fast. such that if then ; wlog, we can take so that we have a legitimate subsequence. Note in particular, . Now, we'll write the telescoping sum in order to use (2): since

(2) tells us that .

Thus, , and we've found a convergent subsequence to .

Remark. If (2) holds, we also have .

Proof of remark. We use the triangle inequality on the partial sums to obtain

and note that LHS and RHS here are convergent sequences of real numbers Reverse triangle tells us , so . so we can take the limit to get the bound on .

So we have that complete normed spaces include infinite sums of elements, if the sum of the norms converges.

Okay , enough meandering. Let be a metric space. Then is a completion of if is isometric An isometry is such that for all , . to a dense subspace of .

Theorem (Cantor Intersection Theorem). Every metric space has a completion which is unique up to an isometry.

If isn't itself complete, this tells us why our Cauchy sequences aren't all converging in : they are "converging'' to a missing point, a point in the completion. If we think about as a subset of (via the isometry), then all the missing points are just limit points of our set : a sequence being Cauchy is equivalent to the sequence being convergent in a larger space.

What is it good for?

Now that we have a good grasp on completeness, we'd like to exploit it to the fullest. Completeness will give us the following:

Let's look briefly at each.

A contraction mapping on a metric space is a map such that for some ,

for all .

Theorem (Banach fixed point). A contraction has a unique fixed point. That is, there exists unique such that .

Proof sketch. If had two fixed points and , the contraction definition would give us a contradiction unless .

To find the fixed point, we choose arbitrarily, then make a sequence . That is, . Since is contracting, if we apply it enough times it stops moving. Since two consecutive terms are exponentially close by contraction:

we can triangle inequality by putting in all the consecutive terms, and estimate with a geometric series to find Cauchyness. Then completeness gives us convergence, and moreover,

since contractions are continuous.

Given a uniformly continuous map with a dense domain, when can we extend it to a uniformly continuous map on the whole metric space? For instance, if is uniformly continuous, when can I extend it to in a way that preserves uniform continuity? This then is simply a question of defining on the limit points of its domain.

Proposition. If is a uniformly continuous map between metric spaces, then maps Cauchy sequences in to Cauchy sequences in .

Proof sketch. One only needs to apply uniform continuity to the definition of a Cauchy sequence. Uniformity is needed to compare the image of any two elements of the sequence far enough along.

This proposition motivates the proof: we use that Cauchyness is defined without needing a limit point so that we can use to map Cauchy sequences in the domain to Cauchy sequences in the codomain without needing to act on a limit point.

Theorem (Banach fixed point). Let and be metric space and suppose is complete. Let be dense, and uniformly continuous. Then there is a unique uniformly continuous extension .

Proof sketch. We use the uniformity of the continuity twice here: once to map Cauchy to Cauchy, and once to get uniformity of our extension out of uniformity of . If we only wanted a continuous extension, we wouldn't need uniformity in the second usage, but we still need uniformity to map Cauchy to Cauchy. So we can't use this method to prove that we can make continuous extensions from continuous functions.

In fact, we cannot always extend continuous functions. The set is dense in , and is continuous, but has no continuous extension (can't define at zero).

If we replace dense with closed and the codomain is , then we can extend continuous functions (this is the Tietze Extension Theorem), but this has nothing to do with completeness, so we won't discuss this further here.
To define on , a limit point of , we choose a sequence in which converges to . This sequence is Cauchy, so converges in . So we define our extension as .

To show this is well-defined, given two sequences and in converging to , we intertwine them to create in converging to ; then exists so its two subsequence and must share the same limit.

To show is uniformly continuous, we consider two limit points and with sequences and in converging to and . We want to show if is close enough to , then is close to . For large , is close to , so by uniform continuity of , is small. But by continuity of the metric,

Uniqueness follows from the uniqueness of limits: any two continuous extensions and must agree on , and by continuity must then agree on limit points.

Onto the Baire Category Theorem (BCT). This theorem is sometimes stated in different but equivalent ways; we'll start with one statement, and then show equivalence to a second.

Theorem (Baire Category Theorem; \Jaksic). We can think of BCT as another way in which complete spaces are not missing bits: open and dense is a strong condition, close to every point and filled in around each point, so if our space isn't missing bits the intersection should still be populated.Let be a complete metric space and be a countable collection of open and dense sets in . Then is dense in .

Proof sketch. We need to show for any open set , is nonempty. We will inductively create a Cauchy sequence which will converge to a point in this intersection, using the density of at each step.

We are exploiting density and openness crucially here: near each point we have something in , and near that something we have enough points to start afresh, and look for a point in .First, since is nonempty since is dense, and open since both sets are, we find some ball . Then we shrink the radius to some to obtain . This ball serves as our "'' in the next step; we intersect with , find some element and a ball around it in the intersection, and repeat. We can take the radius to shrink by at least in each step, and recover a Cauchy sequence with

This tail argument is why we bother messing with the closed balls earlier; we want to guarantee our limit is in the intersection.Our shrinking radii yields that is Cauchy, and completeness gives a limit . Since the tail for each , we have that the limit is in each closed ball too; each , and the first ball is in so the limit .

The theorem fails for non-complete spaces: if we enumerate the rationals as , the sets are open and dense, but the intersection is empty.

Now, we've formulated a statement for open sets; we can take complements, and find

Theorem (BCT corollary). Let be a complete metric space and be closed sets with empty interior. Then has empty interior.

Proof sketch. A set is dense iff has empty interior If is dense, then for , there exists . But then for any ball , there is some , so .

On the other hand, if has empty interior, then for any , , so we can take . Then . For any , the constant sequence . So for any point , we can find a sequence in converging to .
, so taking complements of the , applying BCT, then taking a complement again yields the result.

If we do not decide to take that last complement, we get

Theorem (BCT Drury). Let be a complete metric space and be closed sets with empty interior. Then is dense in .

We then have the remark that for any (nonempty) complete metric space, cannot be written as the countable union of closed sets with empty interior Using the contrapositive of this, since , we have that cannot be complete..

This quickly solves a question: Can we shrink open sets around the rationals to obtain ? No, not if we are shrinking by countable intersections. This concurs with my intuition, but without BCT, I wouldn't know how to prove it.

Proof by BCT. If for sets open in , then we could write

Each of these sets is closed, and we have a countable union. So our last remark to BCT says they cannot all have empty interior. But the have empty interior, so some must have an interior; since is dense, we have . But , so .


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Completeness I - January 1, 2016 - Eric Hanson