Why $H=\bigcup_{n=2}^{\infty}\left[n^{-1}, n^{-1}+2^{-n}\right]$ is a ball closed $G_\delta$ set? I was reading this paper, when in the first page (Definition $1$) I found this claim:

In $\left[0,1\right]$, $H=\bigcup_{n=2}^{\infty}\left[n^{-1}, n^{-1}+2^{-n}\right]$ is a ball closed $G_\delta$ set that is not closed because it does not contain a $0$.

The paper defines a "ball closed $G_\delta$ set" as so:

Definition: A $G_\delta$ subset $H$ of $K$ will be called a ball closed $G_\delta$ set if, whenever $B(x,r) \subseteq H$, $\{y \in K \mid |x-y|=r \} \subseteq H$.

However if I considered the ball $B_{(1/4, 1/4)}$, it is contained in $H$, but its border is not contained in $H$. So $H$ should not be considered ball closed!
Where am I mistaken?
 A: The set $H$ does not contain the ball $B_{(1/4, 1/4)}$.
$$H=\left[{1\over2},{3\over4}\right]\cup\left[{1\over3},{11\over24}\right]\cup\left[{1\over4},{5\over16}\right]\cup\cdots.$$
In particular, $H$ does not contain $\displaystyle{11.5\over24}$.
It took me some time to see this, and it would have been nice if the author had helped out a little more!
A: I'm not convinced that $B(1/4;1/4)$ is contained in $H$.
Consider a visual of the set. I've highlighted the intervals in the union in Desmos:

The areas shaded in purple are the intervals
$$
\left[ \frac 1 n , \frac 1 n + \frac{1}{2^n} \right] \text{, for each $n \in \mathbb{N}$}
$$
Clearly, there are gaps between them all. And, if I choose to overlay $B(1/4;1/4) = (0,1/2)$ in red, we see it falls into those gaps:

You could probably use this idea to motivate a formal proof by picking some $x$ in the gaps and showing it is not in the union, by showing it is not in any of the intervals forming $H$.
A: Here is a general proof as to why $H$ is a ball closed $G_\delta$ set:
Firstly, notice that $K \setminus H =$ {$0, 1$} $\cup \bigcup_{n = 1}^\infty (\frac{1}{n+1} + \frac{1}{2^{n+1}}, \frac{1}{n})$. Suppose that $B(x, r) \subseteq H$, and that $y \in K$ such that $|x - y| = r$. Now, if $y \not\in H$, then we know there exists an integer $n \geq 1$ such that $y \in (\frac{1}{n+1} + \frac{1}{2^{n+1}}, \frac{1}{n})$ (assuming $y \neq 0, 1$, if $y = 0$ then $B(x, r) = B(x, x) = (0, 2x)$ which clearly is not contained in $H$). But it is clear that there exists $z \in (\frac{1}{n+1} + \frac{1}{2^{n+1}}, \frac{1}{n})$ such that $|x - z| < r$ (depending on the choice of $y$, since it can be $x - r$ or $x + r$ provided they are in $K$, one can choose $z > y$ on former case and $z < y$ on latter case). That is to say, $z \in B(x, r)$ but also $z \not\in H$, a contradiction.
Note that I removed the case for when $y = 1$, but that case follows similarly. So, for your example, $B(\frac{1}{4}, \frac{1}{4}) = (0, \frac{1}{2})$, we have that $(\frac{1}{3} + \frac{1}{8}, \frac{1}{2}) \subseteq K \setminus H \cap (0, \frac{1}{2})$ (giving us that $B(\frac{1}{4}, \frac{1}{4})$ is not contained in $H$).
