Metric spaces · Definition

Complete metric space

topologymetric spacescompleteness

Before you start you need to know

Completeness guarantees that a space has no «holes»: every sequence that squeezes together finds its limit inside the space.

Definition

A metric space \(X\) is complete if every Cauchy sequence \((x_n)\) in \(X\) converges to a point \(x\) of \(X\). A sequence is Cauchy if its terms eventually get arbitrarily close to one another.

\[(x_n)\ \text{Cauchy}\ \Longrightarrow\ \exists\,x\in X:\ x_n\to x\]

Formally, \((x_n)\) is Cauchy when for every \(\varepsilon>0\) there is \(N\) such that \(d(x_m,x_n)<\varepsilon\) for all \(m,n\ge N\). Unlike convergence, this condition mentions no limit: it only compares the terms with one another. That is why it can tell, from the inside, whether a sequence ought to converge.

Every convergent sequence is Cauchy, by the triangle inequality, so completeness asks for the converse. The real numbers are the basic example: it is the supremum axiom phrased with sequences. By contrast \(\mathbb{Q}\) is not complete, because the decimal approximations of \(\sqrt2\) are Cauchy and their limit is irrational; neither is \((0,1]\), where \(1/n\) piles up at a \(0\) that does not belong to the space.

Completeness is a property of the distance and not only of the open sets: \(\mathbb{R}\) with \(|x-y|\) and with \(|\arctan x-\arctan y|\) has the same open sets, but only the first is complete. A typical mistake is to think that «bounded» is enough or that completeness depends only on the shape of the set; what matters is that limits of Cauchy sequences do not escape.

Completeness yields powerful results, such as Banach's fixed point theorem and the Baire category theorem. In the next topic we move from distances to the more general notion of a topology.

Think about it

Why does a Cauchy sequence with a convergent subsequence converge?

Because the subsequence terms approach the limit \(x\) and the other terms approach them; the triangle inequality gives \(d(x_n,x)<\varepsilon\) for large \(n\).

Is \([0,1]\) with the usual distance complete?

Yes: a Cauchy sequence converges in \(\mathbb{R}\) and, \([0,1]\) being closed, the limit lies in \([0,1]\).

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §43
  • Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990. cap. 2