My question is about the possibility of the inverse of a well known theorem:
“Let $A_1 \supset A_2 \supset A_3 \supset … $ be nested, non-empty compact sets and let the diameter of $A = \bigcap_n A_n$ be $0$, then the sequence $diam(A_n)$ approaches $0$ as $n \rightarrow + \infty$”
First of all I noticed that this sequence is weakly decreasing, hence it must converge to the infimum of its range, let us call it $\alpha$.
Then it is also clear that $diam(A) \leq \alpha$, which, however doesn’t seem to tell us anything helpful in this case (whereas is the key inequality for the theorem to which I referred at the beginning).
Any help on proving this theorem (if it can be proven at all) is highly appreciated as always!
Best Answer
Suppose $\operatorname{diam}(A_n)$ does not approach zero. Taking a subsequence, if necessary, we may assume there is $r>0$ so that $\operatorname{diam}(A_n) > r$ for all $n$. Then: there exist $x_n, y_n \in A_n$ with $d(x_n,y_n) > r$. Again taking a subsequence, if necessary, we may assume $x_n \to x$ and $y_n \to y$. Of course $x,y \in A$ and $d(x,y) \ge r > 0$. So $\operatorname{diam}(A) \ne 0$.