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?
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
