Conjecture: Is it true that the limiting diameter of a (strictly !) nested family of compact sets must be 0 ??

Consider a family of compact subsets of $$\mathbb{R}^n, C_1 \supset C_2 \supset C_3 \ldots$$.

Also, and this is the important bit,

1) $$C_j$$ has empty interior for all $$j \in \mathbb{N}$$

2) The inclusions are strictly decreasing, that is $$C_i \neq C_j \; \forall i \neq j$$

Can we then prove that $$\lim_{k \rightarrow \infty}\;\mathrm{diam}(C_k) = 0$$ ?

• What if $C_i = \{(a,0,0,0,\ldots,0)\in\mathbb{R}^n\mid |a|\leq 1+\frac{1}{i}\}$? – Arturo Magidin Apr 12 at 19:43
• That works ! Thank you. If you care to write it up as an answer, I can mark it solved. – me10240 Apr 12 at 19:52
• In one dimension, consider the nested closed subsets of $[0,1]$ whose intersection is the Cantor Set. – Arturo Magidin Apr 13 at 5:56
• In $A_j=\{1,0\}\cup \{1/m: j<m\in \Bbb N\}.$ In $\Bbb R$ let $C_j=A_j.$ For $n>1$ let $C_j=A_j\times \{0\}^{n-1}.$ – DanielWainfleet Apr 13 at 18:24

Consider $$\mathbb{R}^m$$,$$m> 1$$,and $$C_n = [0, \frac{n+1}{n}]\times \{0\}^{m-1}$$. For one dimension we can take decreasing Cantor sets in $$[0,1]$$ that all contain $$0$$ and $$1$$ (so have diameter $$1$$).
• You can use those for higher dimensions (append $0$s); for one dimension, I would use the sets whose intersection makes up the Cantor Set. They are all closed and bounded. – Arturo Magidin Apr 13 at 5:57
• Urgh; of course they don’t. But you can probably start with the Cantor set and just remove pieces of it, ending with $\{0,1\}$. – Arturo Magidin Apr 14 at 1:00