Topology: Complete Metric Spaces

[ This article was updated on 8 Mar 13; the universal property is now in terms of Cauchy-continuous maps. ]

On an intuitive level, a complete metric space is one where there are “no gaps”. Formally, we have:

Definition. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges.

Completeness is not a topological property, i.e. one can’t infer whether a metric space is complete just by looking at the underlying topological space. For example, (0, 1) and R are homeomorphic as topological spaces, but the former is not complete (since the sequence (1/n) is Cauchy but doesn’t converge) and the latter is. Clearly, not every subspace of a complete metric space is complete. E.g. R – {0} is not complete since the sequence (1/n) doesn’t converge. However, we have:

Proposition 1. A subset of a complete metric space X is complete if and only if it’s closed in X.

Proof.

Both directions use theorem 1 here. If Y is closed in X, then any Cauchy sequence in Y is also Cauchy in X and hence must converge to some a in X; then a must lie in Y by the theorem.

Conversely, if Y is a non-closed subset of X, then there is a sequence in Y converging to a in X, outside Y. This sequence is Cauchy since it’s convergent in X, but it doesn’t converge in Y; thus Y is not complete. ♦

Proposition 2. If $(X, d_X)$ and $(Y, d_Y)$ are complete metric spaces, then so is $(X\times Y, d)$ where d can be any one of the following metrics:

$((x,y), (x',y')) \mapsto \sqrt{d_X(x,x')^2 + d_Y(y,y')^2}$ ;

$((x,y), (x',y')) \mapsto d_X(x,x') + d_Y(y,y')$ ;

$((x,y), (x',y')) \mapsto \max(d_X(x,x'), d_Y(y,y'))$ .

Proof.

We know (from proposition 3 here) that if a sequence $(x_n, y_n)$ in X × Y is Cauchy, then $(x_n), (y_n)$ are Cauchy in X and Y respectively. Hence, they’re both convergent, say, to $a\in X, b\in Y.$ Thus, $(x_n, y_n) \to (a, b)$ is convergent. ♦

Exercise

Suppose $X_1, X_2, \ldots$ is a countably infinite collection of complete metric spaces. Construct a metric on $X:=\prod_{n=1}^\infty X_n$ as before. Is the resulting metric space complete?

Answer (highlight to read) :

Yes, each projection map π_n : X → X_n is uniformly continuous. So a Cauchy sequence (x_n)₁, (x_n)₂, (x_n)₃, … gives rise to a Cauchy sequence in each X_n. Each of these converges, to say, a_n in X_n. Then by proposition 4 here, we see that (x_n)₁, (x_n)₂, (x_n)₃, … converges to (a_n). ♦

Completion of Metric Space

It turns out there’s a unique way of embedding any metric space into a “smallest” complete metric space.

Definition. A metric space Y is a completion of metric space X if:

X is a metric subspace of Y;

Y is complete; and

X is dense in Y.

Suppose $X\subseteq Y$ satisfies the first two conditions. Then we take $X\subseteq Z:=\text{cl}_Y(X)\subseteq Y.$ Now Z is closed in Y so it is complete by proposition 1 above. Furthermore, X is dense in Z by definition. So $X\subseteq Z$ is now a completion. In other words, the third condition enforces a minimality condition on Y. One can visualise Y as filling up the gaps of X.

Classical example: the completion of Q, under the Euclidean metric, is R.

The key property of the completion is the following.

Universal Property of Completion. Suppose $X\subseteq Y$ is a completion. Then:

for any Cauchy-continuous map $f:X \to Z$ to a complete metric space Z, there is a unique continuous $g:Y\to Z$ such that $g|_X = f.$

Minor Note.

Since g is continuous, it’s automatically Cauchy-continuous here since Y is complete: indeed, if $(y_n)$ is a Cauchy sequence in Y, then it converges to some y, so $(f(y_n))\to f(y)$ is convergent and must also be Cauchy.

Proof.

Uniqueness: if g, g’ satisfy $g|_X = g'|_X$ then since X is dense in Y and Z is Hausdorff, we have g = g’.

Only existence remains: we denote the metric of X and Y by d and that of Z by d’.

Suppose $y\in Y;$ since X is dense in Y, y is a limit of a sequence (x_n) in X. Then (x_n) is Cauchy and since f is Cauchy-continuous, (f(x_n)) is Cauchy in Z. So (f(x_n)) converges to some z and we let g(y) = z.

To show g is well-defined: suppose $(x_n), (x_n') \to y.$ We need to show $(f(x_n)), (f(x_n'))$ have the same limit. It suffices to show $d'(f(x_n), f(x_n')) \to 0$ as n → ∞.

For each ε>0, pick δ>0 such that whenever d(x, y) < δ, we have d’(f(x), f(y)) < ε.
Given this δ, pick N such that when n>N, we have $d(x_n, y)<\frac\delta 2$ and $d(x_n', y)<\frac\delta 2.$
Then when n>N, $d(x_n, x_n') \le d(x_n, y)+d(y, x_n') < \delta \implies d(f(x_n), f(x_n')) < \epsilon.$

Clearly $g|_X = f$ since if $y\in X$ we can pick the constant sequence (y, y, …) in X.

Finally, we need to show g is continuous. By theorem 6 here, it suffices to show that if $(y_n)\to y,$ then $(g(y_n))\to g(y).$ Now since X is dense in Y, we can pick a sequence $(x_n)$ such that $d(x_n, y_n)<\frac 1 n.$ This gives $(x_n)\to y$ also. From our definition of g, we must have $(g(x_n)) = (f(x_n)) \to g(y)$ so we’re done. ♦

Exercise

Prove that if f had been uniformly continuous, so is the induced g.

Answer (highlight to read).

The following replaces the last paragraph in the proof of universal property.

Let ε>0. There exists δ>0 such that whenever x, x’ in X satisfies d(x, x’) < δ, we have d’(f(x), f(x’)) < ε/2. Now suppose y, y’ in Y satisfy d(y, y’) < δ/2.

Pick sequences (x_n), (x_n‘) in X such that (x_n) → y and (x_n‘) → y’.
For large enough n, d(x_n, y) < δ/4 and d(x_n‘, y’) < δ/4 so we get d(x_n, x_n‘) ≤ d(x_n, y) + d(y, y’) + d(y’, x_n‘) < δ.
Thus, for n big, d’(f(x_n), f(x_n‘)) < ε/2.
Since f(x_n) → g(y) and f(x_n‘) → g(y’) by definition, we can pick a large n such that d’(f(x_n), g(y)) < ε/4 and d’(f(x_n‘), g(y’)) < ε/4.
This gives d’(g(y), g(y’)) ≤ d’(g(y), f(x_n)) + d(f(x_n), f(x_n‘)) + d(f(x_n‘), g(y’)) < ε. ♦

Properties of the Completion

With the universal property, there’re a couple of properties we can prove rather easily.

Proposition 3. The completion is unique up to homeomorphism, i.e. if $X\hookrightarrow Y$ and $X\hookrightarrow Y'$ are both completions, then there is a unique homeomorphism f : Y → Y’ which restricts to the identity map on X.

Proof.

Let $i:X\to Y$ and $j:X\to Y'$ be the injections. Since i and j are uniformly continuous, by the universal property (U.P.) on Y, there exists a unique uniformly continuous $f:Y\to Y'$ which extends the identity map id_X on X. Likewise, by the U.P. on Y’, there is a unique uniformly continuous $g:Y'\to Y'$ extending id_X. Composing gives uniformly continuous maps $g\circ f:Y\to Y$ and $f\circ g:Y'\to Y'$ extending id_X. By uniqueness in U.P., both $g\circ f$ and $f\circ g$ must be identity maps. ♦

Thus, we can call Y the completion of X without any ambiguity.

If two metrics are topologically equivalent, the resulting completions can be very different. E.g. (0, 1) and R (under the Euclidean metric) are homeomorphic topological spaces, but the former’s completion is [0, 1] while the latter is already complete. In short, completion is not a topological construction.

Proposition 4. If Y is the completion of X and $X'\subseteq X$ , then the completion of X’ is its closure in Y.

Proof.

Let Y’ = cl_Y(X’). Since Y’ is a closed subset of complete metric space Y, it is complete too. Also, X’ is dense in Y’ by definition, so we’re done. ♦

Proposition 5. If Y₁ is the completion of X₁ and Y₂ is the completion of X₂, then Y₁ × Y₂ is the completion of X₁ × X₂, if we let the metric take on any one of the three possibilities in proposition 2.

Proof.

We already know Y₁ × Y₂ is complete by proposition 2. Also, $\text{cl}(X_1\times X_2) = \text{cl}(X_1)\times \text{cl}(X_2) = Y_1\times Y_2$ so X₁ × X₂ is dense in Y₁ × Y₂. ♦

Proposition 6. If Y₁ is the completion of X₁ and Y₂ is the completion of X₂, then any uniformly continuous map $f:X_1\to X_2$ extends to a uniformly continuous $g:Y_1\to Y_2.$

Proof.

Compose f with $X_2\hookrightarrow Y_2$ to obtain a uniformly continuous map $X_1 \to Y_2$ to a complete space. By universal property of Y₂, this extends to a uniformly continuous $g:Y_1\to Y_2$ such that $g|_{X_1} = f.$ ♦

If f is injective or surjective, the induced g may not be so. For example, consider the map $(0, 1)\to \{z\in \mathbf{C} : |z|=1, z\ne 1\}$ which takes t to exp(2πit). We leave it to the reader to prove its uniform continuity. The induced map on the completion is $[0, 1]\to \{z\in\mathbf{C} : |z|=1\}$ which takes t to exp(2πit) as well, and this is not injective.

For another example, consider the map [1, ∞) → (0, 1] which takes x to 1/x. This is uniformly continuous and induces [1, ∞) → [0, 1], again taking x to 1/x. This map is not surjective. Observe that in both examples, the initial map is a homeomorphism but the inverse, though continuous, is not uniformly continuous.

Existence of Completion

Although we’ve proved many properties of the completion of a metric space, we’ve yet to demonstrate its existence.

Big Theorem. Every metric space (X, d) has a completion (Y, d’).

Let’s prove this step-by-step.

Step 1 : Define Y.

Let Σ be the set of all Cauchy sequences in X. Define a relation on Σ:

For $(u_n), (v_n) \in \Sigma,$ write $(u_n) \sim (v_n)$ if $\lim_{n\to\infty} d(u_n, v_n) = 0.$

It’s not hard to prove that this gives an equivalence relation on Σ (the proof’s left as an exercise, e.g. transitivity follows from the triangular inequality). We define Y to be the set of equivalence classes.

Step 2 : Define d’

If two element $y, y'\in Y$ are represented by Cauchy sequences $(u_n), (v_n)\in \Sigma$ we define the distance function to be:

$d'(y, y') := \lim_{n\to\infty} d(u_n, v_n).$

First we show that the limit exists; since R is complete it suffices to show $r_n := d(u_n, v_n)$ is Cauchy. To do that, note that for all m, n:

$\begin{aligned} d(u_n, v_n) &\le d(u_n, u_m) + d(u_m, v_m) + d(v_m, v_n) \\ \implies r_n - r_m &\le d(u_m, u_n) + d(v_m, v_n).\end{aligned}$

By symmetry, we also have $r_m - r_n \le d(u_m, u_n) + d(v_m, v_n)$ which gives:

$|r_m - r_n| \le d(u_m, u_n) + d(v_m, v_n).$

Since $(u_n), (v_n)$ are Cauchy, for any ε>0, we can find N such that whenever m, n > N, we have $d(u_m, u_n), d(v_m, v_n) < \frac\epsilon 2 \implies |r_m - r_n| < \epsilon.$ Thus $(r_n)$ is Cauchy.

Step 3 : Prove that d’ is well-defined.

Let’s show that d’(y, y’) is independent of our choice of Cauchy sequences. Suppose $(u_n) \sim (u_n')$ and $(v_n) \sim (v_n').$ Arguing as in step 2, we obtain:

$|d(u_n', v_n') - d(u_n, v_n)| \le d(u_n, u_n') + d(v_n, v_n').$

By definition of ~, the two terms $d(u_n, u_n')$ and $d(v_n, v_n')$ on the RHS tends to 0. Hence d’ is well-defined.

Step 4 : Prove that d’ is a metric.

Clearly d’(y, y’) ≥ 0. Suppose d’(y, y’) = 0, and $y, y'\in Y$ are represented by $(u_n), (v_n)\in\Sigma.$ Then $\lim_{n\to\infty} d(u_n, v_n) = 0$ , so $(u_n) \sim (v_n)$ and y = y’. This proves reflexivity. Obviously d’(y, y’) = d’(y’, y).

This leaves the triangular inequality. Suppose $(u_n), (v_n), (w_n)\in\Sigma.$ Then we have $d(u_n, w_n) \le d(u_n, v_n) + d(v_n, w_n)$ for each n. Taking the limit of each term, we get the triangular inequality for d’.

Step 5 : Define an embedding of X as a metric subspace of Y.

Define the map i : X → Y which takes x to the element of Y representing the Cauchy sequence (x, x, x, … ). Clearly d’(i(x), i(x’)) = d(x, x’).

Step 6 : Prove that X is dense in Y.

Let $y\in Y$ be represented by $(x_n)\in\Sigma.$ We claim that (x_n), as a sequence in Y, converges to y; i.e. $\lim_{n\to\infty} d'(x_n, y) = 0.$

For each ε>0, pick N such that when m, n > N, we have $d(x_n, x_m) < \frac\epsilon 2.$ Fixing n, and taking the limit as m→∞, we get $d(x_n, y) \le \frac\epsilon 2 < \epsilon.$ Thus, we’re done.

Step 7 : Prove that Y is complete.

Let $(y_n)$ be a Cauchy sequence in Y. Since X is dense in Y, for each n, pick $x_n \in X$ such that $d'(x_n, y_n) < \frac 1 n.$ We claim that $(x_n)$ is Cauchy: for

$d'(x_n, x_m) \le d'(x_n, y_n) + d'(y_n, y_m) + d'(y_m, x_m) < \frac 1 n + \frac 1 m + d'(y_m, y_n).$

Hence for any ε>0, just pick N such that (i) N > 3/ε and (ii) for any m, n > N, we get $d'(y_m, y_n) < \frac\epsilon 3.$ Thus, when m, n > N, we have $d'(x_m, x_n) < \epsilon$ and $(x_n)$ is a Cauchy sequence in X, which must correspond to some $y\in Y.$

It remains to show $(y_n)\to y.$ Indeed, we have:

$d'(y_n, y) \le d'(y_n, x_n) + d'(x_n, y) < \frac 1 n + d'(x_n, y).$

Step 6 then tells us that $d'(x_n, y)\to 0$ since $(x_n) \to y.$ And we’re done. ♦

10 Responses to Topology: Complete Metric Spaces

Conan says:

April 19, 2015 at 2:06 pm

Thanks for the article – I just have a few questions regarding the universal property of completion. Firstly, we have that if f was uniformly continuous, then g is uniformly continuous. If f was a homeomorphism – would this also mean g is a homeomorphism? How would we go about proving this? Lastly, in your proof of g being uniformly continuous, why do you have d'(y,y’)? Isn’t y and y’ elements of Y? To show g is uniformly continuous, don’t we need to show that for d(y,y’) < delta, then d'(g(y),g(y')) < epsilon?

Thanks for any feedback you can provide.

limsup says:

April 19, 2015 at 3:51 pm

For your first question, I believe the answer is yes, but for a rather trivial reason. The resulting X must be complete, so the completion Y is X. In other words, we’re given $f:X\to Z$ such that

(i) f is Cauchy-continuous;
(ii) f is a homeomorphism
(iii) Z is complete.

Now if $(x_n)$ is Cauchy, then so is $(f(x_n))$ by (i), and this converges to some $z\in Z.$ By (iii), we have z=f(x) for some $x\in X.$ Since $f(x_n) \to f(x)$ and $f^{-1}$ is continuous by (ii), we have $x_n \to x$ .

For your second question, that was a bug, which I’ve now fixed. Thanks! 🙂

- Conan says:
  
  April 19, 2015 at 11:10 pm
  
  Regarding the first part, I’m a bit confused, if you wouldn’t mind explaining. For example, if we had the situation in Proposition 6 above: If $f$ is uniformly continuous, then we can construct a uniformly continuous map $X_{1} \rightarrow Y_{2}$ which would result in a uniformly continuous map in $g$. Suppose $f$ is a homeomorphism between $X_{1}$ and $X_{2}$ – are you saying this must mean that $X_{1}$ and $X_{2}$ are complete? If so – why, I’m struggling to see why this might be the case?
  
  And to follow up with the proof of $g$ being uniformly continuous, I’ve realised I’m still a bit confused when reading through the proof. My understanding is that we need to show that when $d(y,y’) < \delta \implies d'(g(y),g(y')) < \epsilon$, but you seem to have $d(\overline{x_{n}},\overline{x'_{n}}) < \delta \implies d'(g(y),g(y')) < \epsilon$ – could you explain this?
  
  Thanks for your help!
  
  - Conan says:
    
    April 19, 2015 at 11:12 pm
    
    Apologies for the formatting, I thought latex could be used by your reply earlier.
  - limsup says:
    
    April 20, 2015 at 1:01 am
    
    No why must $X_1$ and $X_2$ be complete? If $f:X_1 \to X_2$ is a homeomorphism, then it just extends to a homeomorphism $g:Y_1 \to Y_2$ . The composed map $X_1 \to Y_2$ is not a homeomorphism then. Just take $X_1 = X_2 = \mathbf{Q}$ and $Y_1 = Y_2 = \mathbf{R}$ for instance.
    
    For your second question, I proved that $d(y, y')<\delta/2 \Rightarrow d'(g(y), g(y'))<\epsilon$ .
    
    [ Latex tip for wordpress: use '$' together with latex, e.g. https://en.support.wordpress.com/latex/ ]
Conan says:

April 19, 2015 at 11:22 pm

Sorry, I think I need to clarify my above comment what I am trying to ask is – if f is a homeomorphism – does this mean than g would be a homeomorphism? Thanks.

- limsup says:
  
  April 20, 2015 at 1:06 am
  
  I understood your question and answered it: in my comment I proved that under your assumptions, X has to be complete. Since X is complete, we have Y=X and so if f is a homeomorphism, so is g because g=f.
  
  - Conan says:
    
    April 20, 2015 at 6:44 am
    
    Thanks for the help on the latex.
    
    So to clarify my understanding, what you’re saying is:
    
    If $\hat{X}$ and $\hat{Y}$ are the completions of $X$ and $Y$ respectively, then if $f:X \rightarrow Y$ is a homeomorphism, then $g$ is a homeomorphism. But the composed map say $f':X \rightarrow \hat{Y}$ is not a homeomorphism. (I suppose because then $f'$ is not surjective – but the image of $f'$ would be a homeomorphism right?)
    
    How could we prove that the homeomorphism of $f$ just extends to $g$?
    
    Thanks for your patience.
limsup says:

April 21, 2015 at 1:20 am

Oops, I made a mistake in the reply timed (April 20, 2015 at 1:01 am). It’s not true that if f is a homeomorphism, so is g. In fact, I had explicitly listed some counterexamples above after proposition 6. >_<

- Conan says:
  
  April 21, 2015 at 11:47 am
  
  Hi, in your examples above proposition 6, it seems to say that if $f$ is a homeomorphism, then this does not necessarily mean that $g$ is uniformly continuous. But strictly speaking, homeomorphism doesn’t require uniform continuous does it? It just requires a bijective continuous function, with inverse continuous as well right? If $f: X \rightarrow Y$ is a homeomorphism and uniformly continuous, wouldn’t that mean $g$ is also a homeomorphism? With $g$ extending $f':X \rightarrow \hat{Y}$ where $\hat{Y} is the completion of$ latex Y$? In this case, wouldn’t $g$ be uniformly continuous, and it would also be bijective?