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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?

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  • $\begingroup$ Is the ambient normed linear space finite-dimensional? $\endgroup$
    – fedja
    Commented Apr 2, 2014 at 1:13
  • $\begingroup$ It is $R^n$. i forget to write this. sorry for this . i fixed my error $\endgroup$ Commented Apr 2, 2014 at 1:15

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In Variation et optimisation des formes 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.

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