[ *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 becompleteif 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 and are complete metric spaces, then so is where d can be any one of the following metrics:

- ;
- ;
- .

**Proof**.

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

**Exercise**

Suppose is a countably infinite collection of complete metric spaces. Construct a metric on 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*. Each of these converges, to say,

_{n}*a*in

_{n}*X*. Then by proposition 4 here, we see that (

_{n}*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 acompletionof metric space X if:

- X is a metric subspace of Y;
- Y is complete; and
- X is dense in Y.

Suppose satisfies the first two conditions. Then we take Now *Z* is closed in *Y* so it is complete by proposition 1 above. Furthermore, *X* is dense in *Z* by definition. So 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 is a completion. Then:

- for any Cauchy-continuous map to a complete metric space Z, there is a unique continuous such that

**Minor Note**.

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

**Proof**.

Uniqueness: if *g*, *g’* satisfy 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 since *X* is dense in *Y*, *y* is a limit of a sequence (*x _{n}*) in

*X*. Then (

*x*) is Cauchy and since

_{n}*f*is Cauchy-continuous, (

*f*(

*x*)) is Cauchy in

_{n}*Z*. So (

*f*(

*x*)) converges to some

_{n}*z*and we let

*g*(

*y*) =

*z*.

To show *g* is well-defined: suppose We need to show have the same limit. It suffices to show 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 and - Then when
*n*>*N*,

Clearly since if 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 then Now since *X* is dense in *Y*, we can pick a sequence such that This gives also. From our definition of *g*, we must have 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*‘) in_{n}*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*‘) < δ._{n} - Thus, for
*n*big,*d’*(*f*(*x*),_{n}*f*(*x*‘)) < ε/2._{n} - 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 and are both completions, then there is a unique homeomorphism f : Y → Y’ which restricts to the identity map on X.

**Proof**.

Let and be the injections. Since *i* and *j* are uniformly continuous, by the universal property (U.P.) on *Y*, there exists a unique uniformly continuous which extends the identity map id_{X} on *X*. Likewise, by the U.P. on *Y’*, there is a unique uniformly continuous extending id_{X}. Composing gives uniformly continuous maps and extending id_{X}. By uniqueness in U.P., both and 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 , 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 Y_{1}is the completion of X_{1}and Y_{2}is the completion of X_{2}, then Y_{1}× Y_{2}is the completion of X_{1}× X_{2}, if we let the metric take on any one of the three possibilities in proposition 2.

**Proof**.

We already know *Y*_{1} × *Y*_{2} is complete by proposition 2. Also, so *X*_{1} × *X*_{2} is dense in *Y*_{1} × *Y*_{2}. ♦

Proposition 6. If Y_{1}is the completion of X_{1}and Y_{2}is the completion of X_{2}, then any uniformly continuous map extends to a uniformly continuous

**Proof**.

Compose *f* with to obtain a uniformly continuous map to a complete space. By universal property of *Y*_{2}, this extends to a uniformly continuous such that ♦

If *f* is injective or surjective, the induced *g* may not be so. For example, consider the map which takes *t* to exp(2π*it*). We leave it to the reader to prove its uniform continuity. The induced map on the completion is which takes *t* to exp(2π*it*) 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 write if

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 are represented by Cauchy sequences we define the distance function to be:

First we show that the limit exists; since **R** is complete it suffices to show is Cauchy. To do that, note that for all *m*, *n*:

By symmetry, we also have which gives:

Since are Cauchy, for any ε>0, we can find *N* such that whenever *m*, *n* > *N*, we have Thus 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 and Arguing as in step 2, we obtain:

By definition of ~, the two terms and 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 are represented by Then , so and *y* = *y’*. This proves reflexivity. Obviously *d’*(*y*, *y’*) = *d’*(*y’*, *y*).

This leaves the triangular inequality. Suppose Then we have 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 be represented by We claim that (*x _{n}*), as a sequence in

*Y*, converges to

*y*; i.e.

For each ε>0, pick *N* such that when *m*, *n* > *N*, we have Fixing *n*, and taking the limit as *m*→∞, we get Thus, we’re done.

**Step 7 : Prove that Y is complete**.

Let be a Cauchy sequence in *Y*. Since *X* is dense in *Y*, for each *n*, pick such that We claim that is Cauchy: for

Hence for any ε>0, just pick *N* such that (i) *N* > 3/ε and (ii) for any *m*, *n *> *N*, we get Thus, when *m*, *n* > *N*, we have and is a Cauchy sequence in *X*, which must correspond to some

It remains to show Indeed, we have:

Step 6 then tells us that since And we’re done. ♦

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.

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 such that

(i) f is Cauchy-continuous;

(ii) f is a homeomorphism

(iii) Z is complete.

Now if is Cauchy, then so is by (i), and this converges to some By (iii), we have z=f(x) for some Since and is continuous by (ii), we have .

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

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!

Apologies for the formatting, I thought latex could be used by your reply earlier.

No why must and be complete? If is a homeomorphism, then it just extends to a homeomorphism . The composed map is not a homeomorphism then. Just take and for instance.

For your second question, I proved that .

[ Latex tip for wordpress: use '$' together with latex, e.g. https://en.support.wordpress.com/latex/ ]

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.

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.

Thanks for the help on the latex.

So to clarify my understanding, what you’re saying is:

If and are the completions of and respectively, then if is a homeomorphism, then $g$ is a homeomorphism. But the composed map say is not a homeomorphism. (I suppose because then is not surjective – but the image of would be a homeomorphism right?)

How could we prove that the homeomorphism of $f$ just extends to $g$?

Thanks for your patience.

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. >_<

Hi, in your examples above proposition 6, it seems to say that if is a homeomorphism, then this does not necessarily mean that 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 is a homeomorphism and uniformly continuous, wouldn’t that mean is also a homeomorphism? With extending where latex Y$? In this case, wouldn’t be uniformly continuous, and it would also be bijective?