[ 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 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 πn : X → 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). ♦

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:

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:

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 (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 We need to show have the same limit. It suffices to show as 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. ♦

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’)  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. ♦