Is this a perfect set? Let $\{r_1,r_2,r_3,\ldots\}$ be an enumeration of the rational numbers, and for each
$n\in\mathbb{N}$ set $\epsilon_n=1/2^n$. Define $O=\bigcup_{n=1}^\infty V_{\epsilon_n}(r_n)$ and let $F=O^C$.
Is $F$ a perfect set?
I think the answer is no, so I'm trying to find a point $x\in F$ and $\delta>0$ such that
$$(-\infty, x-\delta]\cup[x+\delta,+\infty)\cup O=\mathbb{R}-\{x\}$$
since, taking complements, this would imply that
$$(x-\delta,x+\delta)\cap F=\{x\}
$$
and thus $x$ is not a limit point of $F$, but I don't know how to find such $x$.
 A: Note that $F$ depends on how you enumerate the rationals - for each enumration $e:\mathbb{N}\rightarrow\mathbb{Q}$, we get an $F_e$. And it is not obvious that the question does not depend on which $e$ you pick! 
I'll show that there is some $e$ whose associated $F_e$ is not perfect. We'll enumerate the rationals in such a way that $\pi$ (for example) is in $F_e$ but is isolated. To do this, we pick a pair of monotonic sequences of rationals: $q_i\rightarrow \pi$ from below, and $p_i\rightarrow\pi$ from above. As long as these sequences converge to $\pi$ "sufficiently quickly" (say, with $p_i-q_i<2^{-{2^i}}$), we can cook up an $e$ such that the open balls $e$ puts around the $q_i$s union to form an open interval whose right endpoint is $\pi$, and the open balls $e$ puts around the $p_i$s union to form an open interval whose left endpoint is $\pi$; and furthermore, $\pi$ is never covered by any of the open balls on any of the points. This ensures that $\pi$ is an isolated point of $F_e$.

I believe, meanwhile, that there are enumerations of $\mathbb{Q}$ which do yield perfect sets, however. Why? Well, pick some enumeration $d$. The associated $F_d$ is an uncountable closed set, so we can write $F_d=P\sqcup C$ where $P$ is the maximal perfect subset of $F_d$ and $C$ is countable. Now, note that no point in $C$ can be a limit point of points in $P$, since the points in $C$ are exactly those points with Cantor-Bendixson rank $<\infty$. But that means we can find an open $U$ containing $C$ with $U\cap P=\emptyset$. 
Indeed, we can find such a $U$ with arbitrarily small measure. So if we just had a little extra measure, we could smoosh things around and cover up $P$ - and this would yield a perfect $F$. But where can we get that extra measure?
Well, in any covering of $\mathbb{Q}$ by open balls, there's lots of overlapping. So it seems likely that we could tweak our enumeration $d$ to get a $d'$ which covers exactly $P^c$. This, however, would take some work to verify, and I'm not certain it's correct.
A: If $x$ is an isolated point of $F$ this implies that $\forall\delta > 0$, $x - \delta \in O$. This means that $\exists n \in N$ such that $r_n + \frac{1}{2^n} = x$ ($x$ is the limit of a neighborhood) which is a contradiction as it requires $x$ to be a rational. As a result, $F$ is a perfect set.
I understand the provided answer but I have no idea which one of the two claims is right...
