Does convergence for Cauchy sequence fail only when the limit is not in the domain?

Question

I am trying to understand how important is the distinction between Cauchy sequences and convergent sequences in normed vector spaces $E$. So far I have only come across examples where the Cauchy sequence $\{x_n\}$ where $x_n\in E$ fails to converge only because the limit point is not in $E$ and an extension to $E$ typically by completion fixes the problem. For example:

$$x_n\colon[0,1]\to\Bbb R, \quad t\mapsto \sum_{k=0}^n\frac{t^k}{k!},$$

where $E\triangleq \mathcal{P}([0,1])$ is the space of polynomial functions on $[0,1]$ with uniform convergence norm. I want to know if this is the only kind of failure mode for the convergence of a Cauchy sequence.

Answer 1

This can be viewed as "the only kind of failure mode", since any non-complete metric space can be viewed as a subspace of a complete metric space. For more details, see e.g. this question and its answers.

Answer 2

Yes, it is the only kind, in the following sense: If $(X,d)$ is a metric space and if $(a_n)_{n\in\mathbb N}$ is a Cauchy sequence of elements of $X$ which does not converge, then there is a metric space $(Y,d^\star)$, such that $Y\supset X$ and that $(\forall x,x'\in X):d^\star(x,x')=d(x,x')$ and the sequence $(a_n)_{n\in\mathbb N}$ converges in $Y$.

Answer 3

A sequence $(x_n)_n$ in a metric space is not convergent iff the set of limit points of the sequence is not a singleton. So iff it is empty or has more than one element.

If $(x_n)_n$ is a Cauchy-sequence then it can be proved that there is no more than one element in that set, so if the sequence is not convergent then the set of limit points must be empty.

This can always be "repaired" by completion.