# Closed nowhere dense positive measure subset of $[0,1]$ without isolated points

I'm trying to prove that there exists a subset $F$ of $[0,1]$ such that:

1. $F$ is closed

2. $F$ is nowhere dense

3. $F$ has a positive Lebesgue measure

4. $F$ contains no isolated points.

I know that Cantor fat sets could be used but I'd like to see a different approach.

I was thinking about using the construction that was described in this thread: link (I mean the construction based on enumerating rationals in $[0,1]$).

Such construction leads to a set which clearly satisfies the first three conditions, but I'm having trouble with the isolated points.

Thank you.

• Are you familiar with the Cantor-Bendixson theorem? Oct 2, 2016 at 10:37
• Unfortunately not. I was hoping for some rather elementary approach. Oct 2, 2016 at 13:43
• The Smith-Volterra-Cantor set is a fairly elementary example. It’s constructed just like the middle-thirds Cantor set, except that the lengths of the open intervals that you remove are smaller and smaller fractions of the lengths of the closed intervals from which you remove them, so that the sum of their lengths is less than $1$. Oct 2, 2016 at 21:08
• @Jenda358: The Cantor-Bendixson theorem is fairly easy to prove, but if you had known it, there would be no need for that. Oct 3, 2016 at 9:30

Let $F$ be any set satisfying conditions 1-3. Enumerate all open balls (or your favourite countable basis of open sets) in the interval as $(B_n)_n$, and put $\mathcal U:=\{B_n\mid B_n\cap F\textrm { is countable}\}$, and let $F':=F\setminus \bigcup \mathcal U$ (this is called the perfect (or Cantor-Bendixson) kernel of $F$). Check that $F'$ satisfies your conditions (note that $\bigcup \mathcal U\cap F$ is a countable open subset of $F$).

This is a special case of a general theorem called the Cantor-Bendixson theorem: any closed subset of a Polish (i.e. separable completely metrisable) space can be written as a disjoint union of a perfect set (i.e. closed, without isolated points) and a countable set.

• Thanks! Very nice and elegant solution! Oct 3, 2016 at 20:55
• @Jenda358: You're welcome. If you like the answers you are given, you should upvote them (by clicking the up arrow on the left), and accept the one you like most (by clicking the check mark under the arrows). Oct 4, 2016 at 7:14

This is a modification of the construction described in the linked answer. The idea is to not take every ball $B(q_n,2^{-(n+1)})$ but to select them in order to avoid isolated points of the complement. To do so, we will take one ball at a time, choosing the radius small enough to our need.

So, enumerate the rationals as $(q_n)_{n\geq 1}$, and let us define inductively on a parameter $k\in\mathbb{N}$ which balls $B_k$ we will take.

[$k=1$] The first step is to take $B_1=B(q_1,\frac14)$.

[$k\to k+1$] Now let $C_k=\bigcup_{i=1}^k B_i$ be the union of all the balls taken up to step $k$. Consider the smallest index $n\in\mathbb N$ such that $q_{n}\in \overline{ C_k}^c$, call it $n(k)$, and take the ball $B_{k+1}=B(q_{n(k)},r_k)$ where $$r_k=\min\left\{\frac {1}{2^{k+1}}, \frac12 dist(q_{n(k)}, \overline{C_k})\right\}.$$ The algorithm is saying: at each step consider the next rational in the enumeration that's not in the closure of the balls already taken, and take the ball centered there of radius $1/2^{k+1}$, unless it falls too close to the balls already taken; in that case, choose a smaller radius. Check that this is all well defined, and let $B=\bigcup_{k=1}^\infty B_k$, and $F=[0,1]\backslash B$.

Now the key observation is that the two extreme points of every ball $B_k$ are contained in $F$ (why?). For every point $q\in F$ there are balls among $B_k$ arbitrarily close to it, and therefore their extremes will have $q$ as a limit. The only way this could not work is if $q$ were itself an extreme of two balls simultanesously. But this is impossible, since the last ball to be chosen between the two could not lie so close to the other.

• Thank you so much! That's the kind of construction I wanted to see! Oct 3, 2016 at 20:45

You already know how to construct $F\subseteq[0,1]$ with the first three properties: closed, nowhere dense, positive measure. But it may have isolated points; you want to know how to get rid of the isolated points.

Let $\mathcal I$ be the set of all rational intervals $I$ such that $I\cap F$ is countable; let $A=\bigcup_{I\in\mathcal I}(I\cap F),$ and let $E=F\setminus A.$ I claim that $E$ has all four properties.

$E$ is closed because $E=F\setminus\bigcup\mathcal I$ and $\bigcup I$ is open.

$E$ is nowhere dense because $E\subseteq F.$

$E$ has positive Lebesgue measure because $F\setminus E=\bigcup_{I\in\mathcal I}(I\cap F)$ is countable.

Suppose $x$ is an isolated point of $E.$ Choose a rational interval $I$ containing $x$ so that $I\cap E=\emptyset.$ Then $I\cap F\subseteq F\setminus E,$ so $I\cap F$ is countable, so $I\in\mathcal I,$ so $x\notin F\setminus\bigcup\mathcal I= E,$ a contradicton. Therefore $E$ has no isolated points.

Take the set $F$ described in the other answer, and then consider the set $$F'=\overline{F^{(1)}}$$ where $F^{(1)}$ is the set of points of $F$ with Lebesgue density $1$.

Now condition 1 is trivial. Condition 2 is satisfied, the new set being smaller. Condition 4 is satisfied because $F^{(1)}$ does not contain isolated points, and taking the closure does not add any. By Lebesgue density theorem $F'$ has the same measure of $F$, thus also condition 3 is satisfied.

• Thank you, but unfortunately I'm not familiar with density a Lebesgue density theorem. Can you think of any more elementary approach? Oct 2, 2016 at 13:47
• @Jenda358 I had thought that maybe you wanted a more elementary approach...I will think about it!
– Del
Oct 2, 2016 at 14:22
• @Jenda358 A way could be the following: replace any isolated point of $F$ (which are at most countably many) with a small Cantor set (or any closed set of measure zero with no isolated points). However this still relies on a Cantor set somehow, and maybe you don't want that
– Del
Oct 2, 2016 at 14:33
• "...any nowhere dense closed set of measure zero..."
– Del
Oct 2, 2016 at 14:41

Thm: If $F\subset [0,1]$ is compact, then $F$ is the disjoint union of a perfect set and a set that is at most countable.

Taking this as known, let $U$ be any open subset of $[0,1]$ containing $\mathbb Q \cap [0,1]$ such that $m(U) < 1.$ Let $F = [0,1]\setminus U.$ Then $F$ is compact, nowhere dense, and has positive measure. Apply the theorem to write $F=F_1 \cup F_2,$ where $F_1$ is perfect and $F_2$ is countable. Then $F_1$ satisfies the requirements 1-4.