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Is there such example that $A\subset \mathbb{R}$, such that both $A$ and $\mathbb{R}\backslash A$ are sets without interior points and with positive Lebesgue measure?

Asaf Karagila
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Sqr
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    Isn't the union of two nowhere dense sets nowhere dense? The existence of such a set would make $\mathbb{R}$ nowhere dense in itself, wouldn't it? – saulspatz Apr 11 '21 at 06:40
  • The question seems to have changed since the answer was written and accepted. It's better to post a new question instead of changing an old one. Anywyay, if by "inner point" you mean "interior point", let $A$ be the union of a fat Cantor set with $\mathbb Q$. –  Apr 11 '21 at 07:40
  • @Bungo Sorry for that, I'll do it later. Do you mean $A = {q+x:x\in E,q\in\mathbb{Q}}$ and $E$ is the fat Cantor set? – Sqr Apr 11 '21 at 07:48
  • No, I mean $E \cup \mathbb Q$ where $E$ is a fat Cantor set. This set has (finite) positive measure and contains no intervals, hence has no interior points. Its complement has infinite measure and contains no rationals hence no intervals. –  Apr 11 '21 at 07:49
  • @Bungo Thank you. – Sqr Apr 11 '21 at 07:51
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    For the new question (I agree that questions with accepted answers shouldn't be changed): take $A=\big(\Bbb Q\cap(0,\infty)\big)\cup\big((-\infty,0)\setminus\Bbb Q\big)$. – Greg Martin Apr 12 '21 at 07:08
  • There is a lot of confusion in the various revisions (I did one) as to whether you want the two sets to have no interior points or you want the two sets to be nowhere dense. The answer is YES either way, and examples are in Greg Martin's comment and Henno Brandsma's answer. For the difference between having no interior points and being nowhere dense, note that "$E$ is nowhere dense" is equivalent to "the closure of $E$ has no interior points". Also, see my answer to Is saying a set is nowhere dense the same as saying a set has no interior? – Dave L. Renfro Apr 13 '21 at 08:58
  • Oops, for "both sets are nowhere dense" the answer is NO; for "both sets have no interior points" the answer is YES. – Dave L. Renfro Apr 13 '21 at 12:50

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You cannot write $\Bbb R$ as a union of two nowhere dense sets, because the union of nowhere dense sets is still nowhere dense.

And without interior points (after the question change): take $A$ negative rationals plus positive irrationals), and the other set its complement.

Henno Brandsma
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