# Bounding the Euclidean distance between a given point and the “boundary” of a manifold

Consider a $$d$$-dimensional smooth manifold $$M$$ in $$\mathbb{R}^D$$ ($$d < D$$). Denote $$B_r(c)$$ as an open Euclidean norm ball in $$\mathbb{R}^D$$ with center $$c$$ and radius $$r$$. For a positive real number $$\Delta < r$$ and $$c \in M$$, let $$K = B_r(c) \setminus B_{r - \Delta}(c)$$ be the region between these two balls. We set $$U = M \bigcap B_r(c) \neq \emptyset$$ and $$V = M \bigcap K \neq \emptyset$$.

Suppose the reach of the manifold is upper bounded by $$\tau$$. We pick $$r < \frac{\tau}{2}$$, and let $$\phi$$ be the projection onto the tangent space at $$c$$. As shown in http://people.cs.uchicago.edu/~niyogi/papersps/NiySmaWeiHom.pdf, the derivative of the projection is nonsingular. Then $$(U, \phi)$$ is a chart for $$M$$.

Let $$x \in V$$. We are interested in bounding the Euclidean distance between $$x$$ and $$\phi^{-1}\left(\overline{\phi(U)}\right)$$, where $$\overline{\phi(U)}$$ denotes the "boundary" of $$\phi(U)$$, i.e., $$\textrm{cl}(\phi(U)) \setminus \textrm{int}(\phi(U))$$ (cl for closure and int for interior).

I am new to the differential geometry and have stuck on this problem for quite a while. My intuition is that since the curvature of the manifold is restricted, by making $$\Delta$$ small, the Euclidean distance can be bounded by a function of $$\Delta$$ and $$r$$. Unfortunately, I cannot come up with any rigorous argument, nor any counterexamples.