Topology of metric completion of Euclidean metric Lets consider $\cal{M}=\mathbb{R}^{2}\backslash\{(0,y)\}\text { with } \{|y|\le1\}$ with the Euclidean metric with line element  $ds^{2}=dx^{2}+dy^{2}$.
Now consider the distance function given by $$d(p,q)=inf\{\int_{\gamma}||\dot{\gamma}||dt\}$$ where the infimum is taken over all $C^{1}$ curves $\gamma:[a,b]\rightarrow \cal{M}$ that satisfies $\gamma(a)=p$ and $\gamma(b)=q$ and $||\dot{\gamma}||$ is the norm of the tangent vector with respect to the euclidean metric.
I want to Cauchy complete this space with respect to that distance function. My initial thought was that the completion should be $\mathbb{R}^{2}$. However, the distance between $(-\frac{1}{n},0)$ and $(\frac{1}{n},0)$ is not zero as $n\rightarrow\infty$. This makes me think that the sequences $\{(-\frac{1}{n},0)\}$ and $\{(-\frac{1}{n},0)\}$ won't be in the same class. 
Is this correct? If the Cauchy completion is not $\mathbb{R}^{2}$ what kind of space it is. What can I say about the topology of the complete space using the topology induced by the metric.  Is it simply connected?   
 A: As you already pointed out, sequences of the type $\left\{\left(-\dfrac{1} {n},y\right)\right\}$ are not equivalent to $\left\{\left(\dfrac{1} {n},y\right)\right\}$, for $|y|\leq1$, so each one will converge to a different point in the complete space. If we call this points $(0^-,y)$ and $(0^+,y)$ their distances will be given by $d=2(1-|y|)$. So the space has something like a "hole" where the line segment cut out of $\Bbb{R}^2$ is. So it is homeomorphic to $\Bbb{R}^2\backslash B(0,1)$, where $B(0,1)$ is the open disc of radius $1$. This can be seen by using the bijective function $f:\mathcal{M}\to\Bbb{R}^2\backslash B(0,1)$, such that
$$f(x,y)=\left\{\begin{array}{ll}(x,y),& \text{for} \, |y|\geq 1 \\
(x+\cos(\arcsin y),y), &\text{for} \,|y|\leq 1 \,\text{and} \,x\geq0^+\\
(x-\cos(\arcsin y),y), &\text{for} \,|y|\leq 1 \,\text{and} \,x\leq0^-\end{array}\right.$$
$f$ maps the old line segment, which now constitutes of two segments connected at $y=\pm1$, to $S^1$ and dislocates the points to the right and to the left. With this construction we can see that the space is not simply connected.
