Topology: Complete Metric Spaces

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

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.

warning 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.


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')).


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. ♦


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 πnX → Xn is uniformly continuous. So a Cauchy sequence (xn)1, (xn)2, (xn)3, … gives rise to a Cauchy sequence in each Xn. Each of these converges, to say, an in Xn. Then by proposition 4 here, we see that (xn)1, (xn)2, (xn)3, … converges to (an). ♦


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.


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

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 Yy is a limit of a sequence (xn) in X. Then (xn) is Cauchy and since f is Cauchy-continuous, (f(xn)) is Cauchy in Z. So (f(xn)) 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(xy) < 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 (yy, …) 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. ♦


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 xx’ in X satisfies d(xx’) < δ, we have d’(f(x), f(x’)) < ε/2. Now suppose y, y’ in Y satisfy d(yy’) < δ/2.

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


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 idX on X. Likewise, by the U.P. on Y’, there is a unique uniformly continuous g:Y'\to Y' extending idX. Composing gives uniformly continuous maps g\circ f:Y\to Y and f\circ g:Y'\to Y' extending idX. 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.

warningIf 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.


Let Y’ = clY(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.


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.


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.

warningIf 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 mn:

\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 mnN, 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’(yy’) 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’(yy’) ≥ 0. Suppose d’(yy’) = 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 yy’. This proves reflexivity. Obviously d’(yy’) = 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 iX → Y which takes x to the element of Y representing the Cauchy sequence (xxx, … ). Clearly d’(i(x), i(x’)) = d(xx’).

Step 6 : Prove that X is dense in Y.

Let y\in Y be represented by (x_n)\in\Sigma. We claim that (xn), 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 mnN, 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 mN, we get d'(y_m, y_n) < \frac\epsilon 3. Thus, when mnN, 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. ♦

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10 Responses to Topology: Complete Metric Spaces

  1. Conan says:

    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.

  2. limsup says:

    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:

      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!

  3. Conan says:

    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:

      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:

        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.

  4. limsup says:

    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:

      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?

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