A space is complete when every sequence that tries to converge succeeds — that is, when every Cauchy sequence has a limit inside the space. Completeness is what separates the real line from the rationals, and it is the property everything that approximates depends on.

A road is complete if every journey that slows to a halt actually stops somewhere on the road. The rationals are a road with pinpricks missing: you can drive towards a hole and never arrive.

Rational approximations of $\sqrt2$ — $1.4$, then $1.41$, then $1.414$ — drawn spread out, since at true scale all three would sit inside the hollow point. Each is a point of $\mathbb{Q}$; the point they close in on is hollow, because it is not one. In $\mathbb{R}$ the same sequence has its limit and the space is complete; in $\mathbb{Q}$ the sequence is Cauchy and converges to nothing.

A metric space is complete when every Cauchy sequence of its points converges to a point of the space. The previous lesson gave the two halves of this: convergent sequences are always Cauchy, and Cauchy sequences are not always convergent. Completeness is the name for spaces where the two classes coincide.

• $\mathbb{R}$ with $d(x,y)=|x-y|$ is complete. This course takes it on trust: the proof needs the least upper bound property, which belongs to the analysis course, and a prerequisite cannot cross a course.
• $\mathbb{Q}$ is not complete, and neither is : each has a Cauchy sequence whose only candidate limit is missing.

The subsequence gets close to the limit and the whole sequence gets close to itself, so any late term can be joined to the limit through a term of the subsequence. Halve the tolerance twice over: once for the Cauchy condition and once for the subsequence, then pick an index of the subsequence lying beyond both. This is the argument that makes sequential compactness worth having later: finding one convergent subsequence is then enough.

Apply the Cauchy condition to half the tolerance.

Apply convergence of the subsequence to the same half.

The indices of a subsequence increase without bound, so one of them lies beyond N as well.

For any n beyond N, travel to the limit through that term of the subsequence.

Every tolerance has been met from an index on, which is convergence of the whole sequence.

Complete means: being Cauchy is already enough to converge, with the limit in the space.

• Complete: every Cauchy sequence converges, with its limit in the space.
• $\mathbb{R}$, $\mathbb{R}^n$, $[0,1]$ and every discrete space are complete; $\mathbb{Q}$ and are not.

Great work! You now know how to:

• define a complete metric space;
• name complete and incomplete spaces and justify each;
• use that a Cauchy sequence with a convergent subsequence converges;
• say which fact about the reals this course takes on trust and why.