A question connected with increasing the dimensionality of Euclidean spaces. [closed]

For each positive integer $n$ let $S(n)$ be a compact convex subset of $n$-dimensional Euclidean space with a non-empty interior and let $S(n)$ be a proper subset of $S(n+1)$. Can anyone provide an example of an infinite sequence $S(1),S(2),...,S(n),...$ of this sort in which, as $n$ approaches infinity, the diameters of these sets remain bounded but their $n$-dimensional volumes (i.e. their $n$-dimensional Lebesgue measures) do not converge to zero?

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Am I missing something here? Suppose all the sets $S(n)$ have diameter at most $d$. Then $S(n) \subset B(x_n,d)$, for some $x_n \in \mathbb{R}^n$, and consequently $$\mathcal{L}^n(S(n)) \le \mathcal{L}^n(B(x_n,d)) = \frac{\pi^{n/2}d^n}{\Gamma(\frac{n}{2}+1)} \to 0$$ as $n \to \infty$. (As you can see this depends on the normalization constants for the measures of the unit balls. If you change those you still see the answer to your question just by looking at the measures of the balls.) Perhaps you are trying to ask something else? – Tapio Rajala Dec 26 2010 at 20:30
to Tapio Rajala: No, you are not missing anything. You have just shown me what should have been obvious to me and why what I was asking for was impossible. It was a stupid question to ask and should probably be deleted. I got confused by trying to understand why the volume of an n-dimensional ball of fixed radius approaches zero as n approaches infinity. – Garabed Gulbenkian Dec 26 2010 at 21:21
Think of the ration between the volumes of the unit sphere and the unit box in n dimensions. for n=1 it is 1/1 For n=2 it is 1/1 in the middle but $\pi$/4 over all. For n=3 it is $\pi$/4 at the equator and would stay $\pi$/4 if we had a cylinder but for the sphere it is $\pi$/6. – Aaron Meyerowitz Dec 26 2010 at 21:38
@Garabed Gulbenkian: OK. Usually I am the one making the stupid mistakes and replies, so it is nice to see it going this way for once. :) And yes, this is probably not a research level question.. but still on the subject: Even if you were given some other Haar measures $\mu^n$, your question would still come down to estimating the measures of the (unit) balls. Either $$\limsup_{n \to \infty}\mu^n(B(0,d)) = \limsup_{n \to \infty}\mu^n(B(0,1))d^n \to 0$$ for all $d >0$ or else you have an example using balls. Therefore there is no need to consider any other convex sets. – Tapio Rajala Dec 26 2010 at 22:13
Voting to close, in view of Tapio's excellent answer. – Gerry Myerson Dec 27 2010 at 2:47