The boundaries of a nested sequence of bounded convex sets converge in the Hausdorff metric Let $\Omega_1 \supset \Omega_2 \supset ...$ a sequence of nonempty, open, bounded and convex sets. Define $\Omega = \operatorname{int} \Bigl( \overline{\displaystyle\bigcap_{k=1}^{\infty} \Omega_k } \ \Bigl) $ and supoose that $\Omega$ is nonempty, open, convex and bounded. My intuition says that $\partial \Omega_k \rightarrow \partial \Omega$ in the Hausdorff distance. I don't know how to prove this. If I prove this  then I understand a theorem. Can someone give me a hint to prove this?
 A: That's true. You could prove that the sequence $\bar \Omega_k$ converges to $\bar \Omega$ in the Hausdorff distance. In fact $\bar \Omega_k$ being bounded have a subsequence converging somewhere. But being decreasing, the whole sequence converges, and the limit is $\bar\Omega$. 
Now you want to prove that $d(\partial \Omega,\partial \Omega_k) \le d(\bar \Omega,\bar \Omega_k)$. To achieve this the nontrivial part is to prove that given $p\in\partial \Omega$ then there exists $q\in \partial \Omega_k$ which is close to $p$. Since $\Omega$ is convex you can find an half line external to $\Omega$ with end-point in $p$. On this line there is some $q\in \partial \Omega_k$. Notice that $d(p,q) = d(q,\bar\Omega) \le d(\bar\Omega_k,\bar\Omega)$.
A: Here is a rough idea: The limit of the boundaries always contains the boundary of the limit, even without convexity. If there is a limit point $x$ of the boundaries of the $\Omega_k$ that is not in $\partial \Omega$, then you can pick another point $y \in \Omega$. If $L$ is the line segment between $x$ and $y$, convexity implies that there are line segments $L_k \subseteq \Omega_k$ such that $L_k \to L$ in Hausdorff distance. Since $x$ is not in the closure of $\Omega$, there is a small disk $D$ about it which is disjoint from $\Omega$. So $\Omega_k \cap D$ has to be contained in a $\epsilon_k$-neighborhood of $L \cap D$, with $\epsilon_k \to 0$. This shows that any line segment from $x$ to a point in $\Omega_k \setminus D$ has to have a slope which is in a $\delta_k$-neighborhood of the slope of $L$, with $\delta_k \to 0$. Passing to the limit, this shows that any line segment in $\Omega$ is parallel to $L$, so that $\Omega$ is contained in a line, contradicting openness.
