Is $|\bar{A}| = 0$ if every point of $A \subset \Bbb{R}$ is isolated from the right?

Question

Consider a subset $A$ of $\Bbb{R}$ such that every points in $A$ is isolated from the right in the following sense

$$ \forall a \in A,\ \exists \epsilon > 0 \ \text{ s.t. } \ A \cap [a, a+\epsilon) = \{ a \}. \tag{*} $$

(Notice that $A$ is at most countable since we can define a map $f : A \to \Bbb{Q}$ such that $f(a) > a$ and $A \cap [a, f(a)] = \{a\}$. This condition forces $f$ to be injective.) Now my question is as follows:

Question. If $\bar{A}$ denotes the closure of $A$ in $\Bbb{R}$, is it true that the Lebesgue measure $|\bar{A}|$ of $\bar{A}$ is always zero?

I believe that the property (*) is close to the discreteness so that it prevents $A$ from fattening when we take the closure to it. I have neither a compelling evidence nor a counter-example, though.

Motivation. Why I consider this question is that I want to write $A^c$ as

$$ A^c = \dot{\cup}_i I_i \cup N $$

for a mutually disjoint intervals $I_i$ and a null-set $N$. If this is true, then for any $a, b \in A$ with $a < b$ I can write $b - a$ as $\sum_{I_i \subset [a, b]} |I_i|$, which is crucial for solving another problem I am considering.

Now notice that any connected component of $A^c$ is either a singleton or a fat interval (interval of either finite or infinite length). Since such fat intervals are at most countable, we may enumerate them as $(I_i)$ and write $A^c$ as

$$ A^c = \dot{\cup}_i I_i \cup \underbrace{ \{ x : \{x\} \text{ is a singleton component of } A^c\}}_{=: N}. $$

Then it is easy to see that $N \subset \bar{A}$ and $\bar{A} \setminus N$ is at most countable. Thus we have

$$ |\bar{A}| = |N|. $$

This explains why my motivation reduces to analyzing $|\bar{A}|$ instead.

Answer 1

$\overline{A}$ can have positive measure.

Let $C\subset [0,1]$ be a fat Cantor set with $0,1\in C$. Write

$$[0,1] \setminus C = \bigcup_{n = 0}^\infty (a_n,b_n)$$

with countably many disjoint open intervals. In each interval $(a_n,b_n)$, define a sequence by

$$x^{(n)}_{k} = b_n - 2^{-1-k}(b_n - a_n)$$

and let

$$A = \{ x^{(n)}_k : n,k \in \mathbb{N}\}.$$

Then every point of $A$ is isolated from the right - $A \cap [x^{(n)}_k, x^{(n)}_k + 2^{-2-k}(b_n-a_n)) = \{x^{(n)}_k\}$ - but $C \subset \overline{A} = A \cup C$, so $\lvert\overline{A}\rvert = \lvert C\rvert > 0$.

It remains to prove that indeed $C\subset \overline{A}$. In the construction of $C$, let $C_m$ denote the compact set at the $m^{\text{th}}$ stage, consisting of $2^m$ compact intervals, each of length $\leqslant 2^{-m}$. Let $x\in C$ and $\varepsilon > 0$. Choose $m > 0$ such that $2^{1-m} < \varepsilon$. $x$ belongs to some interval $J^{(m)}_r = [s,t]$ of $C_m$, and by the choice of $m$, the compact interval $J^{(m-1)}_\rho$ of $C_{m-1}$ containing $x$ is a subset of $(x-\varepsilon, x+\varepsilon)$. At least one of $s$ and $t$ is an endpoint of the open interval $J^{(m-1)}_\rho\setminus C_m$, which is one of the $(a_n,b_n)$, whence $(x-\varepsilon,x+\varepsilon) \cap A \neq \varnothing$. Since $\varepsilon > 0$ was arbitrary, it follows that $x\in \overline{A}$.

Answer 2

Actually the given condition is a bit of a red herring. We can find a set $A$ with the property that for every $a\in A,$ there exists an $\epsilon>0$ such that $(a-\epsilon,a+\epsilon)\cap A = \{a\},$ and $m(\overline A)> 0.$

Let $U$ be an open set containing the rationals with $m(U)<1.$ Then $U$ is open and dense and $m(U^c)=\infty.$ Write $U = \cup I_n,$ where the $I_n$ are pairwise disjoint open intervals. Choose $a_n\in I_n$ and set $A= \{a_1,a_2,\dots \}.$ Then $A$ has the above property.

Note that for each $N,$ $\overline {\cup_{n=1}^N I_n}$ can contain no more than $2N$ points of $U^c.$ Thus, except for a countable set, each $x\in U^c$ is the limit of a sequence of points $x_k \in U,$ where distinct $x_k$'s belongs to distinct $I_{n_k}$'s. This forces $m(I_{n_k})\to 0,$ hence the $a_{n_k}$ are dragged along to converge to $x.$ Thus $m(\overline A) = m(U^c) = \infty.$