# Is there a way to form a sequence of intervals s.t they cover all the rational numbers in $(0,1)$ in a way that $C_n \subset Int(C_{n+1}),$

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.

• What topology are you using? Are the $C_i$ sets of rationals or intervals? – Ross Millikan Jun 26 '18 at 5:24
• 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. – астон вілла олоф мэллбэрг Jun 26 '18 at 5:25
• @астонвіллаолофмэллбэрг See my edit, please. – onurcanbektas Jun 26 '18 at 5:27
• @RossMillikan See my edit, please. – onurcanbektas Jun 26 '18 at 5:27

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 $C_{i+1}\setminus C_i$
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).
• A quibble: this doesn't work well for $n=1$ or $2$ – Ross Millikan Jun 26 '18 at 5:35