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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?

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    For each x it is a sequence of numbers so the usual topology of R.I ws thinking F_n be everywhere 0 exept an interval of lenght that goes to 0 where say f_n =1.But i need at every x an oscillation between 0 and 1.That's why i need this sequence.2011-03-25

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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.

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    @Alexander Thumm Or i can use the final answer without $\ 2^n $ but just $\ n$ in the denominator.In this way the division of $\ [0,2 {\pi} ]$ is more smooth.2011-03-31
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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.

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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.