difficulty in formulation I'm trying to find a sequence of intervals on $\mathbb{R}$ which covers $\mathbb{R}$, the length of the intervals approaches $0$ and each $x$ in $\mathbb{R}$ is inside infinite number of these intervals.
So, I ran into the next idea: take $\{a_n\}_{n\in \mathbb{N}}$ to be enumartion of the rationals. then, for each $m\in \mathbb{N}$ define the sequence $\{b^{(m)}_n\}_{n\in \mathbb{N}}=\{[a_n-1/m,a_n+1/m]\}_{n\in \mathbb{N}}$ and then "concatenate" all these sequences to get a new sequence of intervals and this new sequence holds all the above conditions.
I'm having difficulty to formulate the "concatenation" part.
I think that if I want to construct this new sequence in a formal way then I need to use the fact that a countable union of countable sets is countable but I also need to somehow define an order preserving map from $\mathbb{N}$ to $\mathbb{N}^2$ with the dictionary order.
any help would be much appreciated.
 A: Pick your favorite enumeration $ \left\{q_k \right\}$ of the rationals. Define $$a_n= \left\{I_{n,1},\dots,I_{n,n} \right\}$$ with $I_{nk}$ the interval of length $1/n$ around $q_k$.
The sequence obtained by concatenating $I_{n+1,1}$ after $I_{n,n}$ has the desired properties:


*

*It covers $\mathbb R$ since it contains an interval around each rational.

*The length of the intervals approaches $0$ by construction.

*Each real number is in infinitely many arbitrarily small neighborhoods of rationals, and each rational is contained in infinitely many intervals.

A: Such a map cannot exist, because $\mathbb N$ and $\mathbb N^2$ are not order isomorphic (the latter has $\mathbb N$ many elements with no immediate predecessor). However, there is no need for such a map. Rather than 'concatenating' your sequences, you should 'merge' them together. Let $P = \{ p_n \mid n \in \mathbb N \}$ be an enumeration of all primes. Define a sequence $(s_n \mid n \in \mathbb N)$ by letting $s_{p_m^n} = \frac{b_n^m}{p_m^n}$ (*) and let $s_n = \emptyset$ whenever $n$ is either $1$ or has more than $1$ prime factor. This way you have, for each $m \in \mathbb N$ an infinite sequence $(s_{p_m^n} \mid n \in \mathbb N) = (b_n^m \mid n \in \mathbb N)$ and that's all you need to make your approach work.
(*) For an interval $I = \{ x \in \mathbb R \mid a < x < b \}$, we let $\frac{I}{n} := \{ \frac{x}{n} \in \mathbb R \mid a < x < b \}$.
