An intuitive understanding of Cauchy completeness? The standard procedure for completing a metric space is adding a limit for every Cauchy sequence, thus making the space Cauchy-complete. When elementary analysis is first taught, however, the completeness of the real numbers is usually introduced using the axiom of completeness, which asserts that the real numbers are Dedekind complete, that is, every Dedekind cut is generated by a real number (or equivalently, every upper-bounded nonempty subset has a least upper bound). For the real numbers, the two definitions coincide; but for a general metric space, the Dedekind definition is stronger.
I'm struggling to understand, or to give an intuitive explanation, for why the Cauchy definition truly implies "completeness", in the sense that any "place" (or "hole") in the space will have a point in it. I can rationalize the Dedekind definition: it basically implies that wherever you "cut" the line, you will find a number there; thus there are no "holes". If this definition was provable from the Cauchy one, I would not have complained; However the Cauchy definition is strictly weaker, and thus I'm struggling to see why does Cauchy completeness truly counts as "completeness", in the intuitive or geometric sense of "continuousness". Can anyone find a sort of intuitive or graphical explanation for that?
 A: Maybe it helps to look at a quite simple example of a set that is Dedekind-complete, but not Cauchy-complete.
Take the set $S=(-\infty,0)\cup[1,\infty)\subset\mathbb R$, with the order and metric obtained by restricting the order and metric of $\mathbb R$.
Now intuitively, you see that there's a gaping hole in this set. However, the hole is not visible to the order. Indeed, this set is Dedekind-complete in that every bounded set has a supremum (least upper bound) and an infimum (greatest lower bound).
But wait, you ask, what about the set $N=(-\infty,0)$? Well, it has a supremum in $S$, and that supremum is $1$. That's because $1$ is clearly an upper bound, and in $S$ there's no upper bound that's lower, so $1$ is the supremum of $N$.
In short, the order cannot see the hole in the line; it sees the line as if the two parts were glued together.
On the other hand, the metric does see the hole, after all, all elements of $N$ have a distance larger than $1$ from the element $1$. And in particular, some Cauchy sequences will not converge, for example the sequence $a_n = -1/n$. So this set is not Cauchy complete, and the Cauchy completion will effectively add the $0$ to this set, giving $S'=(-\infty,0]\cup[1,\infty)$.
Interestingly, for $S'$ both the order and the metric will see that there's a gap (the order by seeing that there are two elements, $0$ and $1$, with no element in between), but $S'$ is both Dedekind-complete and Cauchy-complete.
Finally, let's look at the case that both $0$ and $1$ are missing, that is, at the set $S''=(-\infty,0)\cup (1,\infty)$. Now both Dedekind and Cauchy will see that there's a gap (the set $N$ now has no supremum, and the sequence $a_n$ doesn't converge). However Dedekind will fill that gap with a single element (which might represent any single point in the interval $[0,1]$), while Cauchy will fill it with two elements, $0$ and $1$. In other words, it adds the border points of the hole.
So for this case, the Cauchy results more closely resemble the intuitive notions of continuity: If there's a macroscopic gap, it will see it and fill in the border points if missing. While Dedekind might not see the gap at all, and if it does, it will just glue everything together with a single additional point.
