Complete connected manifold implies nonextendible If $M$ is a connected riemannian manifold, we can define it as extendible if there is another connected riemannian manifold $M'$ and there exists a proper open $A \subset M'$ ($A \neq M'$) such that $M$ is isometric to $A$.
Well, let $M$ be a connected riemannian manifold, I have to see that if $M$ is complete, then $M$ is nonextendible. I cannot see how to show it, however I have proved a characterization of completeness, which could be helpful: a riemannian manifold $M$ is complete iff each divergent curve in $M$ has no finite length (a curve $\gamma : [0 , 1) \to M$ is divergent if for each compact $K \subset M$ there exists $t \in [0 , 1)$ such that $\gamma([0 , 1)) \cap K = \emptyset$.
The beginning probably is taking a connected extendible manifold $M$, but how can I take a Cauchy and not convergent sequence $\{x_n\}$ in $M$?
 A: For the proof it suffices to show that if $A \subset M'$ is a proper, nonempty, open subset of a connected Riemannian manifold then $A$ is not complete.
For the proof, pick a point $p \in A$, and a point $q \in M'-A$, and a smooth path $\gamma : [0,1] \to M'$ such that $\gamma(0)=p$ and $\gamma(1)=q$ (the existence of this path follows from a theorem of topology saying that every connected, locally path connected topological space is path connected).
Let ${\cal T}  = \gamma^{-1}(M'-A) = \{t \in [0,1] \mid \gamma(t) \in M'-A\}$. This set ${\cal T}$ is a closed subset of $[0,1]$, because $A$ is open in $M'$ hence $M'-A$ is closed in $M'$. Also ${\cal T}$ is nonempty, because $1 \in \cal T$. Also $0 \not \in T$.
Let $T = \min {\cal T}$, and so $T > 0$. Choose a sequence $t_n \in [0,T)$ that converges to $T$. It follows that $x_n = \gamma(t_n)$ is a sequence in $A$ that converges to $\gamma(T) \in M'-A$. Thus, $x_n$ is a Cauchy sequence in $A$ that has no limit in $A$, so $A$ is not complete.
