Homeotopy of Shrinking Manifolds Let $M$ be a $n$-dimensional open manifold in $\mathbb{R}^n$.  Let $B^n_k$ be the closed $n$-dimensional ball of radius $k$.  Let $$N_k = (M^c \oplus B^n_k)^c$$ where $X^c$ denotes the complement of set $X$ and $A \oplus B$ denotes the Minkowski sum of sets $A$ and $B$.  Then, there exists some $\epsilon>0$ such that $\forall k<\epsilon$, $M$ and $N_k$ are homeomorphic.
The above operation "shrinks" $M$ by $k$ from every exterior point. 
I'm trying to prove this somewhat trivial sounding proposition, but I haven't found the appropriate methodology to apply.  Since $N_k$ is an open manifold, the problem boils down to proving that $N_k$ has the same number of connected components as $M$.
The key to proving this seems to revolve around determining some distance dependent upon the concave regions of $M$, and ensuring that $\epsilon$ is less than that distance, but I'm a still unsure of the best approach to proving something like this.
 A: As stated, your result does not hold; here is a counterexample in $\mathbb{R}^2$.  
Let $C$ be the closed set $\{(\frac{1}{n},0)|n\in\mathbb{N}\}\cup \{(0,0)\}$ and $M=\mathbb{R}^2\setminus C$.  It is an open $2$-manifold in $\mathbb{R}^2$ with a non-finitely generated fundamental group.
Then for all $k>0$, $T=M^c\oplus B^2_k$ is a countable union of closed disks centered at the points of $C$.  However, no matter how small $k$ is, all but finitely many disks intersect in pairs (sorry, I mean in succession - draw the picture and it is clear what I mean) to form a single closed connected set that contains the origin.  As a fun and easy exercise, compute the number of connected components as a function of $k$.  
The complement of this is an open manifold with only finitely many holes (in particular, has a finitely generated fundamental group) so it does not even have the same fundamental group as $M$, let alone be homeomorphic to the latter.
Edit: to see why the result fails even if $C$ and $M$ are connected, look at the previous picture and modify it in the following way:
Let $C'$ be the set of consecutive circles centered at the midpoints of each segment $[\frac{1}{n+1},\frac{1}{n}]$ and with diameter these endpoints.  Note that since the radii go to zero, this set (always with $(0,0)$ appended) is closed.  From each of these circles remove a tiny open arc from the top, much smaller than the radius (if the radius decreases like $\frac{1}{n^2}$, let the size of the arc be $\frac{1}{2^n}$).  The resulting set $C$ is closed (exercise).  Note that $M=\mathbb{R}^2\setminus C$ is homeomorphic to the plane minus the origin, because $C$ is contractible (exercise).  In particular, $M$ is connected.
Now when you apply closed disks of any small radius $k$ to points this curve, the large circles will just be thickened, with no change in topology.  The very small circles will be swallowed up by the disk centered at $(0,0)$ and will still not affect the topology.  But because the gaps in the circle close up very quickly, at some point, long before reaching the disk at zero, a thickened loop will form in one of the intermediate circles.  Therefore, the complement of this thickened set will be disconnected, and therefore not homeomorphic to $M$.
