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It doesn't seem apparent to me why a nowhere dense set should necessarily be the countable union of closed sets. However, I seem to be having trouble constructing a counterexample.

Yankl
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  • @RobertThingum Irrationals aren't nowhere dense – Jakobian Jun 18 '18 at 23:25
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    Every subset of a nowhere dense set is nowhere dense. Inasmuch as there is a nowhere dense set of cardinality $2^{\aleph_0},$ there are $2^{2^{\aleph_0}}$ nowhere dense sets. Since there are only $2^{\aleph_0}$ Borel sets, there must be nowhere dense sets which are not Borel sets. – bof Jun 18 '18 at 23:25
  • Ah ha, excuse me, I completely deleted the word 'nowhere' while reading. – Robert Thingum Jun 18 '18 at 23:26
  • Any set of positive (Lebesgue) measure contains a nonmeasurable subset. Take a nowhere dense set of positive measure (say a fat Cantor set), it will contain a nowhere dense set which is not Lebesgue measurable. – bof Jun 18 '18 at 23:33
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    If you want a constructive counterexample, the set of all two-sided limit points of the Cantor set (i.e. the Cantor set minus the endpoints of the deleted intervals) is not an $F_\sigma$ set; easy exercise using e.g. the Baire category theorem. – bof Jun 18 '18 at 23:46
  • Think about the natural surjection - which is "almost" a bijection - from the Cantor set to $[0,1]$. We can pull back an arbitrary set $X\subseteq[0,1]$ along this to get a nowhere dense set (since the Cantor set is nowhere dense) which is "almost a copy" of $X$ itself. Can you show that if $X$ is sufficiently "complicated," then so is its "near-copy"? – Noah Schweber Jun 19 '18 at 06:14
  • Are you talking about $[0,1]$ or general sets? – dodo Sep 14 '23 at 11:01

1 Answers1

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A counting argument shows the existence of such sets: let $C$ be the middle third Cantor set inside $[0,1]$. This is closed (compact even) and has empty interior, so $C$ is a nowhere dense subset of $\mathbb{R}$. This means that every subset of $C$ is also a nowhere dense subset of $\mathbb{R}$. $C$ has $2^{\aleph_0}$ points (as many as the reals) and so $2^{2^{\aleph_0}}$ many subsets, all of them nowhere dense in $\mathbb{R}$.

It is well known that the number of $F_\sigma$ sets of $\mathbb{R}$, or even the number of Borel sets of $\mathbb{R}$ is $2^{\aleph_0}$ which is much smaller than $2^{2^{\aleph_0}}$, showing that the overwhelming majority of the nowhere dense subsets of $\mathbb{R}$ is not $F_\sigma$.

To construct a concrete example of this: we again go to the Cantor set $C$. Take any countable dense subset $D$ of $C$. Then $C\setminus D$ cannot be $F_\sigma$ because otherwise $C\setminus D= \cup_n F_n$, $F_n$ closed in $C$ (and thus in $\mathbb{R}$). $F_n$ missing $D$ means that all $F_n$ are nowhere dense in $C$ (!) as well. But then $\{\{x\}: x \in D\} \cup \{F_n: n \in \mathbb{N}\}$ are all nowhere dense in $C$ and their union being $C$ contradicts the Baire theorem (for the subspace $C$ which is compact), so this shows that $C \setminus D$ is a nowhere dense in $\mathbb{R}$ (being a subset of $C$) that is not an $F_\sigma$ in $\mathbb{R}$.

Henno Brandsma
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