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?
Sequence of subspaces of $\ [ 0,2{\pi} ]$ with length that goes to $\ 0 $
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0Can you explain a bit more: what did you try? why you think such a sequence exists? Why do you tag it fourier-analysis? – 2011-03-25
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0I think it can solve a problem of Stein,Fourier analysis(a collegue of mine thought it is a nice problem and gave it me-but i dont know fourier analysis!).So with elementary knowledge i was trying to find a sequence of functions that the integral of their squares tend to 0 but the f_n(x) does not converge at no x. – 2011-03-25
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0For your comment, it sounds like a slowly oscillating sequence of functions should work. Although it depends on which topology you have in mind for the last sentence. – 2011-03-25
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0For 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
3 Answers
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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2This doesn't guarantee that every $x \in [0; 2\pi]$ is in some interval $A_n$ for some $n$. – 2011-03-25
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0Thanks for pointing out the mistake. I have corrected it now. – 2011-03-25
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0@Alexander Thumm I edited the question because it was not going to solve my problem(see comments under my question).My mistake – 2011-03-25
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0@t.spero: The new answer also solves your new question. Also note, that these intervals are tightly related to the standard example of the sequence of functions you are trying to construct. – 2011-03-25
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0Btw, you can make my first idea work by restricting to those enumerations $\iota : \mathbb N \to \mathbb Q \cap [0,2\pi]$ satisfying $|\iota(n-1) - \iota(n)| < 1/n$ for all $n \in \mathbb N$. Or more generally $A_n = [\iota(n) - a_n, \iota(n) + a_n]\cap [0,2\pi]$ and $|\iota(n-1) - \iota(n)| < a_n$ for a positive sequence $a_n$ with $a_n \to 0$ as $n \to \infty$ and $\sum_{n\in\mathbb N} a_n = \infty$. – 2011-03-27
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0@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
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.
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.