Let $\Omega\_1 \supset \Omega\_2 \supset \cdots$ be a sequence of nonempty, open, bounded and convex sets in $R^n.$


Define $\Omega = \operatorname{int} \Bigl( \overline{\bigcap\_{k=1}^{\infty} \Omega\_k } \Bigl) $ and suppose that $\Omega$ is nonempty, open, convex and bounded.


I believe that $\partial \Omega\_k \rightarrow \partial \Omega$ in the Hausdorff distance. In an article that I am studying, it appears the author is using this fact. I'm not finding this result in the literature, but it probably is a basic result of topology. Can someone indicate to me a book or an article with a proof or just saying that the affirmation is true?



the validity of a basic statement involving the Hausdorff distance
In *[Variation et optimisation des formes](http://www.springer.com/mathematics/book/978-3-540-26211-4)* by A. Henrot and M. Pierre, in Chapter 2, Exercises section there is the following statement:



> 
> **Exercise 2.12** Let $(\Omega\_n)$ be a sequence of open sets, having the property of $\varepsilon$-cone, which converges to $\Omega$ (in the Hausdorff distance). Then 
>  $$ \overline \Omega\_n \to \overline\Omega \text{ and } \partial \Omega\_n \to \partial \Omega$$
>  in the Hausdorff distance.
> 
> 
> 


Your case is a particularization of this exercise, since every open convex set has an $\varepsilon$-cone property. The hypothesis that $\Omega\_1$ is bounded and $\Omega$ is open is enough to conclude that you can choose $\varepsilon$ uniformly for all $\Omega\_n$ and $\Omega$. 


Maybe this is an overkill, since working with convex, open, decreasing sets is not that hard. I'll see if I can write a proof of the exercise, for this answer to be complete.




---


Let me state briefly the definition of $\varepsilon$-cone property. Denote $C(y,\xi,\varepsilon)$ the open cone centered in $y$, having rotation axis $\xi$, radius $\varepsilon$ and angle at the vertex equal to $\varepsilon$. 



> 
> We say that $\Omega$ has the $\varepsilon$-cone property if for every $x \in \partial \Omega$ there is a direction $\xi\_x$ such that for every $y \in \overline \Omega \cap B(x,\varepsilon)$ we have $C(y,\xi\_x,\varepsilon) \subset \Omega$.
> 
> 
> 


Note that the cone has the same direction, angle and size for a whole neighborhood of a point on the boundary. Open, convex sets $C$ have $\varepsilon$-cone property with $\varepsilon = \min\{L,\arcsin(r/2m)\}$ where


* $L = d(x\_0,\partial C)/2$;
* $m = \sup\_{x \in \partial C}|x-x\_0|$;
* $B(x\_0,r)$ is a ball contained in $C$.


You can see that $\varepsilon$ is bounded from below if $m$ is bounded from above, which is your case. Therefore $\varepsilon$ can be chosen uniform for all your sets and the exercise applies.


