# Motivation behind the definition of complete metric space

What is motivation behind the definition of a complete metric space?

Intuitively,a complete metric is complete if they are no points missing from it.

How does the definition of completeness (in terms of convergence of cauchy sequences) show that?

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Completeness of metric spaces is a generalization of the completeness of the real numbers. It is one of the fundamental properties of the real numbers that all Cauchy sequences converge. (This isn't true, for example, in the rational numbers.) It is equivalent to whatever other form of completeness you may be aware of, e.g. that every bounded set has a least upper bound. – Jim Belk Oct 15 '10 at 2:21

I'm not going to add nothing directly related to your question and previous answers, but make some propaganda of a theorem I like since I was student and which, I believe, says something stronger than comparing some intuitive notion of completness with its definition.

A somewhat related notion of completeness is the geodesical one. The definition may not be too much appealing unless you're interested in differential geometry, but one of its consequences is easy to explain: if a Riemann manifold is geodesically complete, you can join any two points by a length minimizing geodesic. (But geodesic already implies that it minimizes length, doesn't it? Not quite: just locally. So, for instance, the meridian joining the North Pole with London, but going "backward", through the Bering Strait and the Pacific Ocean, then the South Pole, Africa and finally London, is a geodesic, but not a length minimizing one blatantly.)

Anyway, $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ is not geodesically complete, since there is no length minimizing geodesic joining, say, $(-1,0)$ and $(1,0)$, due to the "hole" $(0,0)$. At the same time, as a metric space, $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ is not complete: the Cauchy sequence $(\frac{1}{n}, 0)$ converges to $(0,0)$, but since $(0,0)$ is not in $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ it doesn't have a limit there.

Well, the Hopf-Rinow theorem tells us that this kind of things always happen together: a "hole" for geodesics is the same as a "hole" for Cauchy sequences, since for a (finite-dimensional) Riemann manifold $M$, both notions agree: $M$ is complete as a metric space if and only if it is geodesically complete.

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+1, higher-level than the question but much more intuitive. Was thinking of answering this myself and then saw yours. – Paul VanKoughnett Oct 15 '10 at 7:56
@Paul. It's a nice theorem, isn't it? The result is at the same time strong (putting together a concept coming from analysis -Cauchy sequence- and one coming from differential geometry -geodesic-) and quite intuitive. – a.r. Oct 15 '10 at 8:16
Thanks. Very nice and intuitive way to look at completeness. – Norman Oct 15 '10 at 8:49

To "fill the holes" or "add the missing points" would presumably mean embedding the metric space as a subspace of a larger metric space. To avoid trivialities like placing a line inside the plane, it is required (and it appears to be the only sensible interpretation) that the given space is dense in the larger space: every neighborhood of a point in the larger space contains points of the smaller space.

A complete metric space is one to which nothing new can be added in this way. The "no holes" definition of completeness is then equivalent to completeness defined using Cauchy sequences.

A more precise term than holes would "punctures". Holes also have a topological meaning such as the hole surrounded by an annulus in the plane, the hole in a torus, or the keyhole in a lock.

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but how is the "no holes" definition of completeness equivalent to completeness using Cauchy sequences? – Norman Oct 15 '10 at 2:35
How about an example? Consider the sequence of rational numbers (1, 1.4, 1.41, 1.414, 1.4142, ... ). This is a sequence obtained by successively truncating the decimal expansion of root 2. This sequence is Cauchy in Q, and yet clearly converges to root 2, which is not in Q. So in some sense Q has a "hole" where root 2 ought to be. In R, however, the above sequence converges just fine to root 2. – Bey Oct 15 '10 at 12:21
@Norman: if there are holes, i.e., a dense isometric inclusion of the metric space in a larger metric space, there must be a point P in the larger space not contained in the smaller space. By density, there is a sequence of points (belonging to both spaces) converging to P. It is a Cauchy sequence (in either space) that is convergent in the larger space but not in the smaller space. So a "geometric hole" must also be a "Cauchy hole". – T.. Oct 15 '10 at 14:21

I tend to think of it this way: A Cauchy sequence in any metric space is one such that for all $\varepsilon > 0$ the tail of the sequence is eventually in some $\varepsilon$-ball (not necessarily around the same point for all $\varepsilon$). In other words the only way a Cauchy sequence can fail to converge is if the limit is somehow "not there". Hence the choice of the word "complete" because all of the limits that should be there, are in fact there.

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Failure of a Cauchy sequence to converge is one way that a point could be "not there". There could, potentially, be other ways. Showing that this is not the case is equivalent to the original question. – T.. Oct 14 '10 at 17:24
Yes! this is exactly what I wanted to ask? – Norman Oct 15 '10 at 8:41

When working with metric spaces, we tend to deal with sequences a lot. Completeness is just the condition that those that should converge actually converge.

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But how does the fact that all cauchy sequences should converge in a complete metric space show that there are no "holes" or "points missing" in complete metric space? – Norman Oct 14 '10 at 13:51
The only possible reason for which a sequence that should converge may not converge is that its limit is missing (this is seen by the way completion works: you add the missing limits) – Mariano Suárez-Alvarez Oct 14 '10 at 14:34
@Normal Joel Osment: Having "holes" can be a bit of a misleading intuition. In 2 dimensions, a disk with a finite-sized hole in it can still be a complete metric space. – Rahul Oct 14 '10 at 17:27
@Mariano: If all Cauchy sequences converge then all we know is that there are no "missing Cauchy points" (such as punctures in a surface, or removable discontinuities in a graph). Norman is asking why there can be no other type of missing points. – T.. Oct 14 '10 at 17:41
@T, my answer gives the motivation for isolating the concept of "complete metric space" into a definition: they are interesting because the sequences that we want to converge converge, and we like that. – Mariano Suárez-Alvarez Oct 14 '10 at 17:56

This answer only applies to the order version of completeness rather than the metric version, but I've found it quite a nice way to think about what completeness means intuitively: consider the real numbers. There the completeness property is what guarantees that the space is connected. The rationals can be split into disjoint non-empty open subsets, for example the set of all positive rationals whose squares are greater than two, and its complement, and the reason this works is because, roughly speaking, there is a "hole" in between the two sets which lets you pull them apart. In the reals this is not possible; there are always points at the ends of intervals, so whenever you partition the reals into two non-empty subsets, one of them will always fail to be open.

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For a partially ordered set, there are two separate concepts, gaps (punctures) and jumps. The gaps can be filled by an order-completion process similar to Dedekind cuts. If I remember, the topological meaning is that absence of gaps equals local compactness and absence of jumps is local connectivity. Or something like that. Describing this precisely would make an interesting question on its own. – T.. Oct 14 '10 at 19:55