I was reading (the outline of) a proof a minute ago (Exercise 1.23(a) in Function Theory of Several Complex Variables by Krantz) and this was stated without justification:
Suppose that $\Omega\subset\subset\mathbb{R}^n$ has $C^1$ boundary. Then for each point $p\in\partial\Omega$ there exists a ball $B$ contained in $\Omega$ such that $\partial B\cap\partial\Omega=\{p\}$.
This seems intuitively obvious but I cannot find a way to prove it to my satisfaction.
My first thought was to find a $\varepsilon>0$ that is smaller than the curvature $K(p)$ of $\partial\Omega$ at $p$, then take the ball with radius $1/\varepsilon$ centered at a distance of $1/\varepsilon$ in the inward normal direction of $\partial\Omega$ at $p$. But like I said this isn't satisfactory for me. First, I'm not very familiar with curvature outside of curves and surfaces in $\mathbb{R}^3$. Second, I know the curvature of surfaces in $\mathbb{R}^3$ can be negative, throwing off what was legitimate about my "proof". Third, it was a very hand-wavy proof to begin with.
Does anyone have a more "hands-on" and legimate proof I could work with? Or any hints? Any help is greatly appreciated. Thanks in advance. (I hope the tags are OK, I didn't know how to categorize this.)
To make something like your idea work you need the boundary to be $C^2$ so that it has a well-defined curvature.
With only $C^1$, I believe this claim is in fact false: consider a region that is described near the origin by $y > |x|^{3/2}$ - you should be able to check that this is $C^1$. Then a circle with radius $r$ touching the origin from above is described by $$y=r-\sqrt{r^2 - x^2} = \frac{x^2}{2r}+O(x^3),$$
which will be less than $|x|^{3/2}$ for some small $x$ no matter what $r$ we choose.