Compactness, Completeness, Connectedness, Oh My! (Part II)

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 (X,d) be a metric space. We call a sequence \{x_n\}_{n \in \mathbb{N}} Cauchy if, for any \epsilon > 0, there exists N \in \mathbb{N} so that, for n,m \geq N, we have d(x_n,x_m) < \epsilon.

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 \{x_n\}_{n \in \mathbb{N}} \to x \in X. By the definition of convergence, there exists N \in \mathbb{N} so that for n,m \geq N, we have that d(x_n,x) < \frac{\epsilon}{2} and d(x_m,x) < \frac{\epsilon}{2}. By the triangle inequality, this implies that d(x_n,x_m) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon for all n,m \geq N. 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 f_n: X \to \mathbb{R}, I can prove that this sequence converges (pointwise) simply by showing that |f_n(x) - f_m(x)| < \epsilon. Thus, since \mathbb{R} is complete, this sequence will converge. This is remarkably useful in, say, the theory of normed spaces, since ||\cdot||: V \to \mathbb{R}^+_0 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 d be the Euclidean metric and let d_{*} be the discrete metric. Define X = \{\frac{1}{n}\}_{n \in \mathbb{N}}. Clearly (X,d) is not complete (since it doesn’t contain 0), yet there exists a homeomorphism \alpha: (X,d) \to (X,d_{*}) and one may check that (X,d_{*}) 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 0<\epsilon <1, there is an N so that d(x_n,x_m) < \epsilon but, by definition of the discrete metric, if \epsilon < 1, then x_n = x_m 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 (X,d) be a complete metric space. Let K \subset X. K is complete if and only if K is closed in X.

Proof: Suppose K is closed. Then, given a Cauchy sequence in K, this sequence has a limit \{x_n\}_{n \in \mathbb{N}} \to x because X is complete. Since K is closed, it contains its limit points so that x \in K and therefore K is complete.

Conversely, suppose K is a complete subset of X. Then, we need show that K contains its limit points. Take x \in \bar{K}. By definition of a limit point, there is a sequence converging to x with elements in K. Since a convergent sequence is a Cauchy sequence, this sequence has limit in K by completeness. By uniqueness of limits (since, in fact, metric spaces are Hausdorff), this limit in K must be x. Ergo, x \in \bar{K} implies x \in K so that K is closed. QED.

As a rather simple example, consider \mathbb{R} as a complete metric space. Then, say, the open set (0,1) \subset \mathbb{R} is not complete as a metric space, yet the closed set [0,1] \subset \mathbb{R} 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 (X,d_1) be a complete metric space. If d_1 \sim d_2, then (X,d_2) 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 d_1 \sim d_2, there exists M so that, for any x,y \in X we have d_1(x,y) \leq M d_2(x,y). Thus, a Cauchy sequence with respect to d_2, that is,  d_2(x_n,x_m) < \epsilon is the same as d_1(x_n,x_m) < \frac{\epsilon}{M}. Since (X,d_1) is complete, such a sequence converges and therefore converges in (X,d_2). 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 (X,d) be a metric space. Let A be dense subset of X such that every Cauchy Sequence in A converges in X. Then, X is complete.

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


Let \{x_n\}_{n \in \mathbb{N}} be a Cauchy sequence in X. Thus, there exists N_1 \in \mathbb{N} such that, for r,s \geq N_1, d(x_r,x_s) < \frac{\epsilon}{3}. Since A is dense in X, for each x_n, we may find a_n \in A such that d(x_n,a_n) < 2^{-n}.

Let N_2 \in \mathbb{N} such that, for m \geq N_2, we have 2^{-m} < \frac{\epsilon}{3}. Let N = \max\{N_1,N_2\}. Then, for j,k \geq N, we see (by the triangle inequality and hypotheses) that

d(a_j,a_k) \leq d(a_j,x_j) + d(x_j,x_k)+ d(x_k,a_k) < \epsilon

Thus, \{a_n\}_{n \in \mathbb{N}} is Cauchy and, by supposition, converges to a \in X. We shall show that \{x_n\}_{n \in \mathbb{N}} \to a.

Let N_3 > N be such that, if k \geq N_3, d(a_k,a) < \frac{\epsilon}{3}. Let j > \max\{N,N_3\} be as above. Then,

d(x_j,a) \leq d(x_j,a_j) + d(a_j,a) < \epsilon

Therefore, \{x_n\}_{n \in \mathbb{N}} \to a and so X 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: A completion of a metric space (X,d) is a complete metric space (X^{*}, d^{*}) along with an isometry (distance-preserving function) \varphi: X \to X^{*} so that \varphi(X) is dense in X^{*}.

For example, the completion of (0,1) is simply [0,1]. 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 \mathscr{C}(X), 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 (X,d) be a metric space. Fix p \in X and define a mapping

\varphi: X \ni q \to f_q \in \mathscr{C}(X) by \varphi: q \mapsto f_q = d(x,q)-d(x,p)

This function is continuous as it is the difference of two continuous functions (see Basic Continuity) and is bounded by the reverse triangle inequality |d(x,p)-d(x,q)| \leq d(p,q). Therefore, f_q \in \mathscr{C}(X).

Moreover, \varphi is an isometry, since, given q_1,q_2 \in X, we have

\sup \limits_{x \in X} |f_{q_1}(x) - f_{q_2}(x)| = \sup \limits_{x \in X} |d(x,q_1) - d(x,q_2)| = d(q_1,q_2).

Ergo, taking the completion of the image of \varphi(X) we obtain a closed subset of \mathscr{C}(X), 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 (X,d) be a metric space.

Call two Cauchy sequences \{p_n\}_{n \in \mathbb{N}} and \{q_n\}_{n \in \mathbb{N}} in X equivalent if \lim \limits_{n \to \infty} d(p_n,q_n) = 0.. I leave it to the reader to check that this indeed satisfies the axioms of an equivalence relation.

Let X^{*} be the set of all equivalence classes so obtained. If P \in X^{*}, and Q \in X^{*}, \{p_n\}_{n \in \mathbb{N}} \in P and \{q_n\}_{n \in \mathbb{N}} \in Q, define

\Delta(P,Q) = \lim \limits_{n \to \infty} d(p_n,q_n).

To see that \Delta defines a metric on X^{*}, we need show that \Delta(P,Q) is unchanged if \{p_n\}_{n \in \mathbb{N}} and \{q_n\}_{n \in \mathbb{N}} are replaced by equivalent sequences.

Let X^{*} = X/\sim. Given P,Q \in X^{*}, let \{p_n\}_{n \in \mathbb{N}}, \{r_n\}_{n \in \mathbb{N}} \in P and \{q_n\}_{n \in \mathbb{N}}, \{s_n\}_{n \in \mathbb{N}} \in Q. We have \Delta(P,Q) = \lim \limits_{n \to \infty} d(p_n,q_n).
Note that

\lim \limits_{n \to \infty} d(r_n,s_n) \leq \lim \limits_{n \to \infty} d(r_n,p_n) + \lim \limits_{n \to \infty} d(p_n,q_n) + \lim \limits_{n \to \infty} d(q_n,s_n)

so the right hand side is just equal to \lim \limits_{n \to \infty} d(p_n,q_n) by the definition of the equivalence. Do the same for the other elements in P and Q to see the two inequalities

\lim \limits_{n \to \infty} d(r_n,s_n) \leq \lim \limits_{n \to \infty} d(p_n,q_n) and \lim \limits_{n \to \infty} d(p_n,q_n) \leq \lim \limits_{n \to \infty} d(r_n,s_n)

Combining these inequalities, we get that \Delta(P,Q) = \lim \limits_{n \to \infty} d(p_n,q_n) = \lim \limits_{n \to \infty} d(r_n,s_n) so that \Delta is a well-defined metric on X^{*}.

For each p \in X, there is a Cauchy sequence all of whose terms are p. Let P_p be the element of X^{*} which contains this sequence. We need show that \Delta(P_p, P_q) = d(p,q) for all p,q \in X. In other words, the mapping \varphi(p) = P_p is an isometry \varphi: X \to X^{*}.

This is simple. Let P_p, P_q \in \varphi(X). Then, \Delta (P_p,P_q) = \lim \limits_{n \to \infty} d(p_n,q_n) = d(p,q). Thus, \psi: X \to X/\sim 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 \varphi(X) is dense in X^{*}, and that every Cauchy sequence in (\varphi(X),\Delta) converges in (X^{*},\Delta).

We need show that, given any element P \in X^{*}, and any \epsilon >0, there is an element P_x \in \varphi(X) such that \Delta(P,P_x) < \epsilon. Choose \{x_n\}_{n \in \mathbb{N}} \in P. Since this is Cauchy, choose N \in \mathbb{N}, such that, if m,n \geq N, d(x_n, x_m) < \epsilon. Apply \varphi to x_m so that \varphi(x_m) = P_{x_m}. Then, \Delta(P,P_{x_m}) = \lim \limits_{k \to \infty} d(x_k,x_{m_k}) < \epsilon. Thus, \varphi(X) is dense in X^{*}.

We now apply Theorem 3 to \varphi(X) \stackrel{dense}{\subset} X^{*}.

Thus, we need find a Cauchy sequence in \varphi(X) that converges in X^{*}. Let \{x_n\}_{n \in \mathbb{N}} be a Cauchy Sequence in X. Then, given \epsilon > 0, there is N \in \mathbb{N} such that, for n,m \geq N, we have d(x_n,x_m) < \epsilon. For each x_j, let \varphi(x_j) = P_{x_j}. Then \{P_{x_n}\}_{n \in \mathbb{N}} is Cauchy, since \varphi is an isometry. Let \varphi(x_m) = P_{x_m}. Again, since \varphi is an isometry, \Delta(P_{x_n},P_{x_m}) = \lim \limits_{k \to \infty} d(x_{n_k},x_{m_k}) < \epsilon. Thus, \{P_{x_n}\}_{n \in \mathbb{N}} \to P_{x_m} \in X^{*}. Ergo, by the Theorem 3, X^{*} 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 (X,d) 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 X to show that this converges and so that the intersection is dense.

Proof: Let \{V_n\}_{n \in \mathbb{N}} be a countable collection of open dense subsets. Let W be an arbitrary open subset in X. We need show that W \cap \bigcup \limits_{n \in \mathbb{N}} V_n \neq \emptyset. Consider W \cap V_1. This is an open set as the intersection of two open sets, so there exists 0< \delta_1 <1 and x_1 so that C_{\delta_1} (x_1) \subset (W \cap V_1) for the closed ball C.

Now, assume that \delta_{n-1} < \frac{1}{n-1} and x_{n-1} have been chosen, then, V_n \cap B_{\delta_{n-1}}(x_{n-1}) is again open by density of V_n, so we may choose \delta_n < \frac{1}{n} so that C_{\delta_n} (x_n) \subset (V_n \cap B_{\delta_{n-1}} (x_{n-1})). Thus, given any \epsilon > 0, choose \frac{1}{N} < \frac{\epsilon}{2}. Then, for n,m \geq N, we see that x_n,x_m \in B_{\delta_N}(x_N). So, by the triangle inequality,

d(x_n,x_m) \leq d(x_n, x_N) + d(x_N, x_m) < \epsilon.

We therefore conclude that \{x_n\}_{n \in \mathbb{N}} is Cauchy, so it converges by the completeness of X. You can check that this means x \in V_n for each n. Finally, we conclude that x \in \Big(W \cap \bigcap \limits_{n \in \mathbb{N}} V_n \Big) or that the intersection is indeed dense, since W 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


J.T. (S.D.)


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