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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 $\lim_{k \rightarrow \infty}\;\mathrm{diam}(C_k) = 0$ ?

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

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  • $\begingroup$ These sets don’t have empty interior. $\endgroup$ – Arturo Magidin Apr 13 at 5:55
  • $\begingroup$ @ArturoMagidin you’re right. I misread as non-empty. $\endgroup$ – Henno Brandsma Apr 13 at 5:56
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 13 at 5:57
  • $\begingroup$ @ArturoMagidin The Cantor set intermediate stages do not have empty interior either. $\endgroup$ – Henno Brandsma Apr 13 at 20:42
  • $\begingroup$ Urgh; of course they don’t. But you can probably start with the Cantor set and just remove pieces of it, ending with $\{0,1\}$. $\endgroup$ – Arturo Magidin Apr 14 at 1:00

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