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2 added counter-example to the second question

Added. Here is a counter-example to the second question.

Let $X=(\mathbb R,d)$ from the above example. Define $Y=(\mathbb R,d')$ similarly by setting $d'(x,y)=2^{-n}|x-y|$ for $x,y\in[n,n+1]$. Then $Y$ is isometric to $[0,1)$. Consider the disjoint union of $X$ and $Y$. For $t\in [0,+\infty)$, denote by $t_X$ and $t_Y$ the copies of $t$ in $X$ and $Y$. For every $t$, attach an arc $\gamma_t$ of length $\ell_t:=10\cdot 2^{-t}$ connecting $t_X$ to $t_Y$. Let $Z$ denote the union of $X$, $Y$ and all these arcs. We have yet defined distances on $X$, on $Y$ and on every arc $\gamma_t$. The metric on $Z$ is defined as the maximal metric bounded above by these metrics on these subsets. It is easy to see that $X$ and $Y$ are embedded into $Z$ isometrically, as well as sufficiently short intervals of the arcs $\gamma_t$ (e.g. their halves are sufficiently short).

The counter-example is the space $Z\setminus Y$. Its completion is $Z\cup\{+\infty\}$. The space cannot be contracted to $+\infty$ for the same reasons as above: the arcs $\gamma_t$ added to $X$ form a tree and don't help to go around weird parts of $X$.

On the other hand, $Z\setminus Y$ can easily be contracted to any point of $Y$ because $\gamma_t$ nearby $t_Y$ is isometric to a straight line segment $(0,\epsilon_t)$. Just map every point of $Z\setminus X$ to a point in this segment in such a way that the distance to $t_Y$ gets multiplied by a small positive constant (like $\epsilon_t/100$).

Thus adding "contraction points" to $Z\setminus Y$ yields $Z$. But $Z$ can be contracted to $+\infty$ - just project everything to $Y$ and contract there.

1

There is a "contraction space" which is not complete. For example, consider a metric $d$ on $[1,+\infty)$ such that for $x,y\in[n,n+1]$ where $n\in\mathbb N$ one has $d(x,y)=2^{-n}|x-y|^{1/n}$ (other distances are defined by gluing the segments together). The completion is obtained by adding one point at $+\infty$.

But there is no map contracting this space to $+\infty$. Indeed, any Lipschitz map $f$ with $f(x)>n+1$ for some $x\in[n,n+1]$ must be a constant on $[n,n+1]$, so there will be a finite fixed point.