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Is Smith-Volterra-Cantor set[https://en.m.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set ] only example of positive Lebesgue measure compact set which does not contain interval?

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closed as off-topic by user21820, Lord_Farin, Holo, Did, Namaste Dec 24 '18 at 12:00

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    $\begingroup$ how do you define "only"? The union of the usual Cantor middle-third set, with a fat Cantor set (or what you call Smith-Volterra-Cantor set) would also have these properties. $\endgroup$ – Mirko May 20 '18 at 4:08
  • $\begingroup$ Yeah. I mean are there another type (not union of compact set and fat cantor set) of positive measure compact set containing no interval? $\endgroup$ – Santanu Debnath May 20 '18 at 4:35
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    $\begingroup$ a compact set of reals would have a smallest point $a$ and a largest point $b$, and infinitely many open intervals in between that are removed, so in the end it does not contain any interval. It might contain some isolated points, and it must be uncountable, and hence contain a homeomorphic copy of the usual Cantor set. More or less, it would be the only example, subject to clarifying the meaning of "only", and taking some possible but perhaps inessential variations into account. Or,perhaps I should say it is not the only one, since in general it need not itself be homeomorphic to a Cantor set $\endgroup$ – Mirko May 20 '18 at 4:41
  • $\begingroup$ Thanks @Mirko. Your comment is good enough for me. $\endgroup$ – Santanu Debnath May 20 '18 at 4:57
  • $\begingroup$ Related --- Nowhere dense subsets of $[0,1]$ with positive measure other than fat Cantor sets $\endgroup$ – Dave L. Renfro May 20 '18 at 16:00
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$(1).$ Example. Let $S=\{s_n: n\in \Bbb N\}$ be any countable dense subset of $(0,1).$ Let $(t_n)_{n\in \Bbb N}$ be any sequence in $\Bbb R^+$ such that $\sum_{n\in \Bbb N}t_n<1/2.$

Let $U= \{(-t_n+s_n,\;t_n+s_n)\cap (0,1):n\in \Bbb N\}.$ Let $V=[0,1]\backslash \bigcup U.$ Since the measure of $\bigcup U$ cannot exceed $2\sum_{n\in \Bbb N}t_n ,$ which is less than $1,$ the measure of $V$ is not $0.$

And $V$ has empty interior because $[0,1]\backslash V\supset S$ and $S$ is dense in $[0,1].$

$(2).$ In general, if $V$ is compact with empty interior and positive measure let $a=\min V$ and $b=\max V .$ Then $(a,b)\backslash V=\bigcup F$ where $F$ is a countable family of pair-wise disjoint open sub-intervals of $(a,b).$ With $m$ denoting Lebesgue measure, we have $\sum_{f\in F}m(f)=m(\bigcup F) =(b-a)-m(V)<b-a.$

Note: In $(1)$, the family $U$ is not necessarily pair-wise disjoint, but $\bigcup U=\bigcup F$ where $F$ is a countable family of pair-wise disjoint open sub-intervals of $(0,1).$

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  • $\begingroup$ If $L=\lbrace (-t_m+s_n, t_m+s_n)\cap (0, 1): m,n \in \mathbb{N} \rbrace $ then is $\cup U=\cup L$? $\endgroup$ – Santanu Debnath May 21 '18 at 2:36
  • $\begingroup$ No. Let $T=\sup_{m\in N} t_m.$ For each $n$ we have $(-T+s_n,T+s_n) \cap (0,1)\subset \cup L.$ Since $T>0$ and $S$ is dense in $(0,1)$ this implies $\cup L=(0,1).$ $\endgroup$ – DanielWainfleet May 21 '18 at 21:54

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