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Is there a way to form a sequence of intervals $C_i \subset \mathbb{R} $ such that they cover all the rational numbers in $(0,1)$ in a way that $$C_n \subset Int(C_{n+1}),$$ where $C_i$ is closed.

Note: It is the standard topology.

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  • $\begingroup$ What topology are you using? Are the $C_i$ sets of rationals or intervals? $\endgroup$ – Ross Millikan Jun 26 '18 at 5:24
  • $\begingroup$ You insist that $C_i$ be a partition, so they must be disjoint, but you also insist on some containment relation. One has to give way. $\endgroup$ – астон вілла олоф мэллбэрг Jun 26 '18 at 5:25
  • $\begingroup$ @астонвіллаолофмэллбэрг See my edit, please. $\endgroup$ – onurcanbektas Jun 26 '18 at 5:27
  • $\begingroup$ @RossMillikan See my edit, please. $\endgroup$ – onurcanbektas Jun 26 '18 at 5:27
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Now you just need a set of nested closed intervals with union $(0,1)$. You can just have $ C_i=[\frac 1{i+3},1-\frac 1{i+3}]$ for example. Then to form a partition of the rationals just take the rationals in $C_{i+1}\setminus C_i$

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Yes: $$C_n = \left[\frac{1}{n},1-\frac{1}{n}\right].$$

This answers the question in the body, not the question in the title - the $C_n$ do not form a partition, since they are not disjoint (indeed, the condition $C_n\subset Int(C_{n+1})$ implies they cannot be disjoint).

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    $\begingroup$ A quibble: this doesn't work well for $n=1$ or $2$ $\endgroup$ – Ross Millikan Jun 26 '18 at 5:35

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