Is my solution to the following exercise ((9) from chapter 4 of Curves and Surfaces, 2nd edition, by Montiel and Ros) correct?:
Let $S$ be a compact connected surface and let $\Omega$ be its inner domain. Prove that if $x \notin \overline \Omega$ then there are at least two lines tangent to $S$ passing through $x$.
Thanks in advance and kind regards.
Solution:
Consider the support function of $S$ based on $x$: $$ f(p) = \langle p - x, N(p) \rangle, \quad p \in S. $$ Since $S$ is compact, $f$ attains both a maximum and a minimum on $S$. Note that the points through which there is a tangent line passing through $x$ are precisely those $p$ such that $f(p) = 0$.
Let $p_m$ and $p_M$ be the closest and farthest points of $S$ with respect to $x$, respectively. It is easy to see that $f(p_m) > 0$ and $f(p_M) < 0$.
Take the pencil of planes whose axis is the segment joining $p_m$ and $p_M$. Since $S$ is compact, for at least one of these planes the intersection with $S$ is a compact, connected, closed curve. Now, if we walk along this curve, we must go though at least two points at which $f$ equals zero. Indeed, if we start at $p_m$, when we reach $p_M$ the function $f$ will have changed its sign. The same happens when we complete the turn, going from $p_M$ to $p_m$. The conclusion then follows.
I think that the given answer is already flawless. But I will add some : We define $S_+ =\{ p\in S| f(p)>0\} ,\ S_-=\{ p\in S| f(p)<0\}$ From the above, $S_\pm$ are nonempty sets.
For $p\in S_+$, let $f(p)=\epsilon >0$. Here $f^{-1} (\frac{\epsilon }{2},\frac{3\epsilon}{2} )$ is open in $S$ and $f$ has positive values on this set. Hence $S_\pm$ are open in the closed surface $S$. Hence $B=\partial S_- =\partial S_+$ is not a point set.