Given a line $L$ in $\mathbb{R}^2$ and a point $p$ in $L$ there is a line $L^{\perp}$ perpendicular to $L$ at $p$ where for every $q$ in $L^{\perp}$ the point in $L$ closest to $q$ is always $p$. Can this be extended to smooth manifolds with smooth boundary?
Suppose I have a metrically complete smooth manifold $M$ with smooth boundary consisting of closed path components. Pick $p$ in the boundary of $M$ and let $(\varphi, U)$ be a chart around $p$. Take $\epsilon > 0$ small so that $B(p,\epsilon) \subset U$. Does there exist $q$ in $B(p,\epsilon)$ such that for $a(q):= d(p,q)$ we have $B(q,a(q)) \cap \partial M$ is empty? Or put alternatively, is there some $q$ in $B(p,\epsilon)$ such that $d(p,q) = d(q,\partial M)$? That is, does there always exist a $q$ for which $p$ is the closest element of $\partial M$(perhaps not uniquely as in the case of the line)? This seems to be the case if $M$ is Riemannian, but it is unclear to me if it is simply a metric space with a smooth structure.
My basic idea was to look over an expression something like
$$\inf_{s \in \bar{B}(q,a(q) - \frac{1}{n})} |d(s,p) - d(s,\partial M)|$$
for some fixed $q$ in $U \cap M$ and some $n$ very large, but this was unsuccessful. Or then maybe take the $\inf$ again over some compact collection of $q$ contained in $int(M)$, but the choice of such a collection is non-obvious to me. Thanks much for any help.
Since you do not assume any relation between the metric and the smooth structure, the answer is negative: Just take $M=\{(x,y): x\ge 0, y\ge 0\}$ with the Euclidean metric and equip it with the smooth structure given by a homeomorphism $M\to \{(x,y): y\ge 0\}$. Then the point $p=(0,0)\in M$ will not be the nearest point to any $q \in M$.
On the other hand, if $M$ is a Riemannian manifold then the answer is positive: It follows from considering the normal exponential map to $\partial M$: For each $p\in \partial M$ consider the geodesic $\gamma_p$ emanating from $p$ and meeting $\partial M$ orthogonally.