Sequence of subspaces of $\ [ 0,2{\pi} ]$ with length that goes to $\ 0 $  Can you give me an example of a sequence of subspaces of $\ [0,2{\pi}] $ that the legth of them tends to $\ 0 $ but for every $\ x \in [0,2{\pi}]$ there are infinitely many of them such as  x lives in them but also infinitely many of them such that x doesn't live in them?
 A: Let $\iota : \mathbb N \to \mathbb Q \cap [0,2\pi]$  be an enumeration. Now take $A_n = [\iota(n) - 1/n, \iota(n) + 1/n] \cap [0,2\pi]$.
As pointed out by chandok the previous answer was wrong.
$$B_{n,k} = [2\pi (k-1) 2^{-n}, 2\pi (k +1) 2^{-n}] \cap [0,2\pi]$$ for $0\leq k \leq n$ and use $ B_{0,0}, B_{0,1}, B_{1,0}, B_{1,1}, B_{1,2}, B_{2,0}, \dots $ as a solution.
A: Without extra assumptions this seems quite easy.  Take the Cantor construction but do not throw anything away, and fatten each interval at each subdivision by $\epsilon/2^n$ (except endpoints).  The resulting collection is uncountable, and I have no idea why you really want it, but there you go.
A: You want the measures of the subsets to converge to $0$, but the sum of the measures to diverge.
Take any sequence of positive reals whose terms go to $0$ such that the sum diverges. Let this be $\{a_i\}$. Let $p_i$ be the partial sum $p_i=\sum_{n=0}^i a_n$. Then consider the sequence of subsets $[p_i (\text{mod} 2\pi), p_{i+1} (\text{mod} 2\pi))$ on the circle where we identify $0$ and $2\pi$. The number of times $x$ is covered by the intervals up to $[p_{i-1},p_i)$ is about $p_i/2\pi$, and since this goes to $\infty$, every $x$ is covered infinitely often.
