As the title says, I would like to know how to construct a compact subset of $\mathbb{R}$ which does not have interior points but positive Lebesgue-measure?
As a hint, it is given $(0,1)\setminus K$...
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what is an inner point? do you mean interior point? – GA316 Nov 06 '13 at 17:48
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Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set. – Nov 06 '13 at 17:51
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1http://en.wikipedia.org/wiki/Nowhere_dense_set#Nowhere_dense_sets_with_positive_measure this may help you. – GA316 Nov 06 '13 at 17:53
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what is $k$ in the hint? – GA316 Nov 06 '13 at 17:54
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Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.
In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $\frac{a}3$ of each component interval at each stage, rather than the entire middle third.
Cameron Buie
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Could you please have a look on this: http://classes.yale.edu/fractals/labs/paperfoldinglab/fatcantorset.html Why does this set have no interior points? – Nov 06 '13 at 18:05
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Well, notice what happens at each stage. In the first stage, we remove an interval of length $\frac14,$ leaving us two intervals of length $\frac38.$ In the second stage, we remove an interval of length $\frac16,$ leaving us $4$ intervals of length $\frac5{32}$. In the third stage, we remove $4$ intervals of length $\frac1{64},$ leaving us $8$ intervals of length $\frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $\frac1{256},$ leaving $16$ intervals of length $\frac{17}{512}$. And so on.... – Cameron Buie Nov 06 '13 at 18:18
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1More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $\dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$\lim_{n\to\infty}\frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $\ell,$ there is some stage at which no interval of length $\ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $\ell,$ then the fat Cantor set contains no open interval. – Cameron Buie Nov 06 '13 at 18:21