$|\{x \in O \mid \operatorname{dist}(x, \mathbb{R} \setminus O) < \varepsilon\}| \geq C \varepsilon$ for all $0<\varepsilon < \delta$

Question

Let $O \subset \mathbb{R}$ be a nonempty open set.

Does there exist $C, \delta >0$ such that $|\{x \in O \mid \operatorname{dist}(x, \mathbb{R} \setminus O) < \varepsilon\}| \geq C \varepsilon$ for all $0<\varepsilon < \delta$ ?

Answer 1

Let $x_0\in O$. Now, since $O\subsetneq\mathbb{R}$ we have $y\in\mathbb{R}\setminus O$. Suppose wlog $y>x_0$ and take $I:=\left[x_0,x_1\right[$ where $x_1=\sup\left\{x\in O\mid\left[x_0,x\right[\subseteq O\right\}$. Now, let $\delta:=d(x_0,x_1)$ and $d_\varepsilon(O):=\{x\in O\mid d(x,\mathbb{R}\setminus O)<\varepsilon\}$ for all $\varepsilon>0$. This is the moment to note that $(x_1-\varepsilon,x_1)\subseteq d_\varepsilon(O)$ provided that $\delta>\varepsilon>0$. Hence, $\varepsilon=\lambda(\left]x_1-\varepsilon,x_1\right[)\leq\lambda(d_\varepsilon(O))$

Answer 2

I guess that, by $\left| A\right|$ you mean the measure $\lambda(A)$ of the set (if you intend the cardinality of $A$ the question make no sense to me). Well, I think the question is about boundedness. Let $O=\bigcup_{n\in\mathbb{Z}}\left]n-\frac{1}{4},n+\frac{1}{4}\right[$. Now, for every choice of $\varepsilon\in\mathbb{R}^+$ we have that your set, namely $d_\varepsilon(O):=\left\{x\in O\mid d(x,\mathbb{R}\setminus O)<\varepsilon\right\}$ (a sort of approximation of $\partial{O}$), has measure $\lambda(d_\varepsilon(O))=+\infty$. In fact it happens to be a (countable) union of open intervals all of the same - positive - size.

Answer 3

If I take just any open interval $I \subset O$ (which can be done since $O$ is open), and if I set $C=2$ and $\delta = \frac{|I|}{2}$, then your statement will hold (?)