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 \(x\), then all terms are eventually close to the limiting valueUsing the triangle inequality, \(d(x,x_n) \leq d(x, x_{n_k}) + d(x_{n_k}, x_n)\)
. This is an equivalence of course: if a sequence is convergent, then it is both Cauchy and has a convergent subsequence.
A metric space \(X\) 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 \(X\) 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, \(X\) and \(Y\) are complete spaces iff \(X\times Y\) is complete, since both Cauchyness and convergence just pass to the components. More interestingly, are subspaces of complete metric spaces also complete?
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 \((x_n)\) that has each element in \(X\) is Cauchy in \(Y\) iff it is Cauchy in \(X\). The connection to closedness is that to be closed means to contain all of the limit points. So we start by considering \(X\) as a subset of \(Y\), and looking at its limit points and thus sequences which converge in \(Y\) but whose elements are contained in \(X\). Then since Cauchyness is equivalent to convergence in \(Y\), we pass to Cauchy sequences, which then lets us consider sequences in \(X\). The proof is short, and feels a little magical to me.
\[\begin{aligned} \operatorname{cl}X \subseteq X &\iff \forall\, (x_n)\,\text{convergent in }Y\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_Y x_n \in X \\&\iff \forall\, (x_n)\,\text{Cauchy in }Y\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_Y x_n \in X \\&\iff \forall\, (x_n)\,\text{Cauchy in }X\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_X x_n \text{ exists} \\&\iff X \text{ is complete.} \end{aligned}\]
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 \(A_1\supseteq A_2\supseteq A_3\supseteq \ldots\) is contracting if \[ \lim_{n\to\infty} \sup \{ d(x,y): x,y \in A_n\} =0. \]
This leads us to a theorem:
Theorem 1 (Cantor Intersection Theorem) Given a metric space \((X,d)\), the following are equivalent:
\(X\) is complete.
For any contracting sequence \(F_n\) of non-empty closed subsets of \(X\), there exists \(x\in X\) such that \[ \bigcap_{n=1}^\infty F_n = \{x\}. \]
(Sketch). To show \((1)\Rightarrow(2)\), we simply take a sequence \(x_n \in F_n\) 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 \(F_n\). A final application of the contracting condition gives us uniqueness.
On the other hand, given a Cauchy sequence \((x_n)\), we can define a contracting\((x_n)\) being Cauchy gives us the contracting condition.
sequence of closed sets by \(F_n = \operatorname{cl}\{x_k: k\geq n\}\); then (2) gives us convergence.
There are more ways complete spaces are determined by missing bits. Let’s upgrade to a normed space.
Proposition 2 Let \((V,\|\cdot\|)\) be a normed space. Then the following are equivalent:
\(V\) is complete.
If \(v_j\in V\) for \(j=1,2,\dotsc\) are such that \(\sum_{j=1}^\infty \|v_j\| \lt \infty\), then there exists \(v\in V\) with Recall that \(\lim_{n\to\infty} \sum_{j=1}^n v_j = v\) means \(\lim_{n\to\infty} \| \sum_{j=1}^n v_j - v\| = 0\).
\[ \lim_{n\to\infty} \sum_{j=1}^n v_j = v. \]
(Sketch). To show \((1)\Rightarrow(2)\), we just note that the partial sums \(s_n = \sum_{j=1}^n v_j\) form a Cauchy sequenceusing the absolute convergence of the norms
, and hence converge to some \(v\in V\).
For \((2)\Rightarrow(1)\), consider a Cauchy sequence \((x_n)\) in \(V\). By our earlier remarks, we only need to find a convergent subsequence. Choose \(n_k\) 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 \(m,n\geq n_k\) then \(\|x_m - x_n\| \lt 2^{-k}\); wlog, we can take \(n_k \gt n_{k-1}\) so that we have a legitimate subsequence. Note in particular, \(\|x_{n_{k+1}} - x_{n_k}\| \lt 2^{-k}\). Now, we’ll write the telescoping sum \(x_{n_k} - x_{n_1} = \sum_{k=1}^{n-1} x_{n_k+1} - x_{n_k}\) in order to use (2): since
\[\begin{aligned} \sum_{k=1}^{\infty} \|x_{n_k+1} - x_{n_k}\| \leq \sum_{k=1}^\infty 2^{-k} = 1 \lt \infty,\end{aligned}\]
- tells us that \(x_{n_k} - x_{n_1}=\sum_{k=1}^{n-1} x_{n_k+1} - x_{n_k} \xrightarrow{V} x\in V\).
Thus, \(x_{n_k} \to x - x_{n_1} \in V\), and we’ve found a convergent subsequence to \((x_n)\).
If (2) holds, we also have \(\|v\| \leq \sum_{j=1}^\infty \|v_j\|\).
(Proof of remark). We use the triangle inequality on the partial sums \(s_n = \sum_{j=1}^n v_j\) to obtain
\[\begin{aligned} \|s_n\| \leq \sum_{j=1}^n \|v_j\|.\end{aligned}\]
and note that LHS and RHS here are convergent sequences of real numbersReverse triangle tells us \(| \|s_n\| - \|v\| | \leq \|s_n - v\| \to 0\), so \(\|s_n\| \xrightarrow{\mathbb{R}} \|v\|\).
so we can take the limit \(n\to \infty\) to get the bound on \(\|v\|\).
So we have that complete normed spaces include infinite sums of elements, if the sum of the norms converges.
Okay, enough meandering. Let \((X,d_X)\) be a metric space. Then \((Y,d_Y)\) is a completion of \((X,d_X)\) if \((X,d_X)\) is isometric An isometry is \(f:X\to Y\) such that for all \(x,y\in X\), \(d_Y(f(x),f(y)) = d_X(x,y)\).
to a dense subspace of \((Y,d_Y)\).
If \(X\) isn’t itself complete, this tells us why our Cauchy sequences aren’t all converging in \(X\): they are "converging’’ to a missing point, a point in the completion. If we think about \(X\) as a subset of \(Y\) (via the isometry), then all the missing points are just limit points of our set \(X\): 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:
- Contraction mappings on complete spaces have a unique fixed point
- Uniformly continuous functions with complete codomainsthe space you’re mapping into
can be extended. - Countable intersections of open and dense sets are dense in complete spaces.This is the Baire-Category Theorem.
The third point can be reformulated as the countable union of closed sets with empty interior has empty interiorI might say this as "In a complete space, you can’t make an interior by a countable union.’’
, and also leads to two important results: the open mapping theorem and the Banach-Steinhaus theorem, each of which may be discussed in future posts.
Let’s look briefly at each.
A contraction mapping on a metric space \((X,d)\) is a map \(f: X\to X\) such that for some \(0 \leq \alpha \lt 1\),
\[ d(f(x), f(y)) \leq \alpha d(x,y) \]
for all \(x,y \in X\).
(Sketch). If \(f\) had two fixed points \(f(x) = x\) and \(f(y) = y\), the contraction definition would give us a contradiction unless \(x=y\).
To find the fixed point, we choose \(x_0\in X\) arbitrarily, then make a sequence \(x_n = f(x_{n-1})\). That is, \(x_n = \underbrace{f\circ f \circ \dotsm \circ f}_{n\text{ times}} (x_0)\). Since \(f\) is contracting, if we apply it enough times it stops moving.
Since two consecutive terms are exponentially close by contraction:
\[\begin{aligned} d(x_{n+1},x_n) = d(f(x_n), f(x_{n-1}))\leq \alpha d(x_n, x_{n-1})\leq \alpha^n d(x_1,x_0),\end{aligned}\]
we can triangle inequality \(d(x_n,x_m)\) by putting in all the consecutive terms, and estimate with a geometric series \(\alpha^m\sum_{k=1}^\infty \alpha^k = \frac{\alpha^m}{1-\alpha}\) to find Cauchyness. Then completeness gives us convergence, and moreover,
\[\begin{aligned} \lim x_{n} = \lim x_{n+1} = \lim f(x_n) = f( \lim x_n)\end{aligned}\]
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? This then is simply a question of defining \(f\) on the limit points of its domainFor instance, if \(f: (0,1) \subset \mathbb{R}\to \mathbb{R}\) is uniformly continuous, when can I extend it to \(\tilde{f}: [0,1]\to \mathbb{R}\) in a way that preserves uniform continuity?
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(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 \(f\) to map Cauchy sequences in the domain to Cauchy sequences in the codomain without needing to act \(f\) on a limit point.
(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 \(f\). 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 functionsIf we replace dense with closed and the codomain is \([-1,1]\), 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.
. For example, the set \(A = (0,1)\subset \mathbb{R}\) is dense in \([0,1]\), and \(f: A \to [-1,1],\, x \mapsto \sin(1/x)\) is continuous, but has no continuous extension (can’t define at zero).
To define \(\tilde{f}\) on \(x\), a limit point of \(A\), we choose a sequence \((x_n)\) in \(A\) which converges to \(x\). This sequence is Cauchy, so \(f(x_n)\) converges in \(Y\). So we define our extension as \(\tilde{f}(x) = \lim_Y f(x_n)\).
To show this is well-defined, given two sequences \((x_n)\) and \((y_n)\) in \(A\) converging to \(x\), we intertwine them to create \((z_n)\) in \(A\) converging to \(x\); then \(\lim f(z_n)\) exists so its two subsequence \(f(x_n)\) and \(f(y_n)\) must share the same limit.
To show \(\tilde{f}\) is uniformly continuous, we consider two limit points \(x\) and \(y\) with sequences \((x_n)\) and \((y_n)\) in \(A\) converging to \(x\) and \(y\). We want to show if \(x\) is close enough to \(y\), then \(\tilde{f}(x)\) is close to \(\tilde{f}(y)\). For large \(n\), \(x_n\) is close to \(y_n\), so by uniform continuity of \(f\), \(d_Y(f(x_n),f(y_n))\) is small. But by continuity of the metric,
\[ d(f(x_n),f(y_n)) \xrightarrow{n\to\infty} d(\tilde{f}(x), \tilde{f}(y)). \]
Uniqueness follows from the uniqueness of limits: any two continuous extensions \(h\) and \(g\) must agree on \(A\), 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.
Let \(X\) be a complete metric space and \(\{O_n\}_{n=1}^\infty\) be a countable collection of open and dense sets in \(X\). Then \(\bigcap_{n=1}^\infty O_n\) is dense in \(X\).
(Sketch). We need to show for any open set \(O\), \(O\cap \bigcap_{n=1}^\infty O_n\) is nonempty. We will inductively create a Cauchy sequence \((x_n)\) which will converge to a point in this intersection, using the density of \(O_j\) at each step.
We are exploiting density and openness crucially here: near each point we have something in \(O_n\), and near that something we have enough points to start afresh, and look for a point in \(O_{n+1}\).
First, since \(O\cap O_1\) is nonempty since \(O_1\) is dense, and open since both sets are, we find some ball \(B(x_1,\epsilon) \subset O\cap O_1\). Then we shrink the radius to some \(r_1 \lt \epsilon\) to obtain \(\operatorname{cl}B(x_1,r_1) \subset O\cap O_1\). This ball \(B(x_1,r_1)\) serves as our "\(O\)’’ in the next step; we intersect with \(O_2\), find some element and a ball around it in the intersection, and repeat. We can take the radius to shrink by at least \(r_n \lt 1/n\) in each step, and recover a Cauchy sequence \((x_n)\) with
\[ x_n \in \operatorname{cl}(B(x_n,r_n)) \subset O_n \cap B(x_{n-1},r_{n-1}). \]
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 \(r_n \lt \frac{1}{n}\) yields that \((x_n)\) is Cauchy, and completeness gives a limit \(x = \lim x_n\). Since the tail \(\{ x_n: n\geq k\} \subset \operatorname{cl}(B(x_k,r_k))\) for each \(k\), we have that the limit is in each closed ball too; each \(\operatorname{cl}B(x_k,r_k) \subset O_k\), and the first ball is in \(O\) so the limit \(x\in O\cap \bigcap_{n=1}^\infty O_n\).
The theorem fails for non-complete spaces: if we enumerate the rationals as \(\mathbb{Q}= \{r_1,r_2,\ldots\}\), the sets \(\mathbb{Q}\setminus \{r_n\}\) are open and dense, but the intersection is empty.
Now, we’ve formulated a statement for open sets; we can take complements, and find
(Sketch). A set \(A\) is dense iff \(A^c\) has empty interior. Why?
- If \(A\) is dense, then for \(x\in A^c\), there exists \(A\ni x_n \to x\). But then for any ball \(B(x,\epsilon)\), there is some \(x_n \in B(x,\epsilon)\), so \(B(x,\epsilon) \not \subseteq A^c\).
- On the other hand, if \(A\) has empty interior, then for any \(x\in A\), \(B(x,1/n) \not \subseteq A\), so we can take \(x_n \in B(x,1/n) \cap A^c\). Then \(x_n\to x\). For any \(x\in A^c\), the constant sequence \(x\to x\). So for any point \(x\in X = A\cup A^c\), we can find a sequence in \(A^c\) converging to \(x\).
Using this, by taking complements of the \(F_n\), applying BCT, then taking a complement again we obtain the result.
If we do not decide to take that last complement, we get
We then have the remark that for \(X\) any (nonempty) complete metric space, \(X\) cannot be written as the countable union of closed sets with empty interiorUsing the contrapositive of this, since \(\mathbb{Q}= \bigcup_{n=1}^\infty \{r_n\}\), we have that \(\mathbb{Q}\) cannot be complete.
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This quickly solves a question: Can we shrink open sets around the rationals to obtain \(\mathbb{Q}\)? No, not if we are shrinking by countable intersections. This agrees with my intuition, but without BCT, I wouldn’t know how to prove it.
(Proof by BCT). If \(\mathbb{Q}= \bigcap_{j=1}^\infty U_j\) for sets \(U_j\) open in \(\mathbb{R}\), then we could write
\[\begin{aligned} \mathbb{R}= \mathbb{Q}\cup \mathbb{Q}^c = \bigcup_{n=1}^\infty \{r_n\} \cup \bigcup_{j=1}^\infty U_j^c.\end{aligned}\]
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 \(\{r_n\}\) have empty interior, so some \(U_j^c\) must have an interior; since \(\mathbb{Q}\) is dense, we have \(\mathbb{Q}\cap U_j^c\neq \emptyset\). But \(U_j^c \subseteq \mathbb{Q}^c\), so \(\mathbb{Q}\cap U_j^c = \emptyset\).