Completeness for bimetric spaces Let $(X,d)$ be a complete metric space. Is it possible to find a second metric $d'$ such that $d(x,y) \le d'(x,y)$, $\forall x,y\in X$ for which $(X, d')$ is not complete?  
 A: All finite metric spaces are complete, so it is never possible to find such a metric on finite metric spaces.
If $(X,\,d)$ is a complete discrete metric space, it is also not possible, since a converging sequence in a discrete space is finally constant, and $d' \geqslant d$ implies that all $d'$-Cauchy sequences are also $d$-Cauchy sequences, hence finally constant, and convergent.
If the topology may change, it is possible to find such a metric $d'$ for all non-discrete complete metric spaces. Pick any $d$-accumulation point $a \in X$ and define
$$d'(x,y) = \begin{cases} 1 + d(x,y),\quad x \neq y \land (x = a \lor y = a)\\
d(x,y)\quad\;\;,\quad \text{otherwise.}
\end{cases}
$$
$d' \geqslant d$ is clear, as is $d'(x,y) = 0 \iff x = y$. The symmetry is also clear, it remains to check the triangle inequality $d'(x,z) \leqslant d'(x,y) + d'(y,z)$.
That is fulfilled if none of $x,\,y,\,z$ is $a$ by the triangle inequality for $d$. If $x = z = a$, it is fulfilled by the non-negativity. If $x \neq y = a \neq z$, $d'(x,z) = d(x,z) \leqslant d(x,y) + d(y,z) = d'(x,y) + d'(y,z) - 2$. If exactly one of $x$ and $z$ - without loss of generality $x$ is $a$, we have $d'(x,z) = 1 + d(x,z) \leqslant 1 + d(x,y) + d(y,z) = d'(x,y) + d'(y,z)$ (which distance absorbs the $1$ depends on whether $y = a$).
Since $a$ was supposed to be an accumulation point, there is at least one sequence $(x_n)$ in $X \setminus \lbrace a\rbrace$ converging to $a$, and this is a $d'$-Cauchy sequence, since on $X \setminus \lbrace a\rbrace$ the two metrics coincide. Since the topology of $(X,\,d')$ is finer than that of $(X,\,d)$, the only candidate for a limit of $(x_n)$ is $a$, but by the definition of $d'$ and $x_n \in X \setminus \lbrace a \rbrace$, we have $d'(a,x_n) \geqslant 1$ for all $n$. Thus there is at least one Cauchy sequence in $(X,\,d')$ that doesn't converge.
Now, the question remains whether it is possible to have a complete metric space $(X,\,d)$ and a metric $d'$ on $X$ that dominates $d$ and induces the same topology, but such that $(X,\,d')$ is not complete.
That is not possible.
Let $(X,\,d)$ be a complete metric space, and $d' \geqslant d$ a metric on $X$ that induces the same topology as $d$.
Then $(X,\, d')$ is a complete metric space.
Let $(x_n)$ be a $d'$-Cauchy sequence in $X$. Since $d' \geqslant d$, it is also a $d$-Cauchy sequence. Hence, since $(X,\,d)$ is complete, there is a $\xi \in X$ such that $x_n \to \xi$ (convergence is a purely topological and not a metric property).
