Is it possible for a Cauchy sequence to have no limit point at all? So I'm studying completeness of metric spaces, and it seems that most incomplete spaces are subsets of another space which contains the limits of their Cauchy sequences. I was wondering if it were possible for an incomplete metric space to have a Cauchy sequence that has no limit in any space whatsoever, including any space that could contain that incomplete space. Thanks.
 A: It so happens that every metric space $(X,d_X)$ has a completion, $(\hat{X},d_{\hat{X}})$ which comes equipped with an embedding $i:X\hookrightarrow{} \hat{X}$ so that $d_{\hat{X}}(i(x),i(x'))=d_X(x,x').$ The construction essentially "formally" adds the limits of Cauchy sequences by defining the elements of $\hat{X}$ to be Cauchy sequences $(x_n)$ of elements of $X$ modulo the equivalence relation stating that $(x_n)\sim (y_n)$ if and only if $\lim_{n\to\infty} d_X(x_n,y_n)=0$. The elements of $X$ can be identified with the equivalence classes of the constant sequences $(x_n=x)$. 
Actually, applying this construction to Cauchy sequences with respect to the usual norm on $\mathbb{Q}$ gives us the field $\mathbb{R}$. Applying this construction to $\mathbb{Q}$ with the $p-$adic norm $\lvert \:\cdot\:\rvert_p$ produces the $p-$adic numbers $\mathbb{Q}_p$.
A: Good thinking. But you should go one step further: given an incomplete space, if you don't know whether it is contained in a larger metric space, why not make one yourself?
The process is called completion. To put it simply, you see every Cauchy sequence as a single object, and define the equivalence classes of Cauchy sequences. All these equivalence classes constitue a new space that is complete. The original space can be seen as a subspace of (embedded into) the new space.
Since you can always do completion. The answer to you question is: there's always a larger space that's complete.
