Is $[\star]$ a limit point of the sequence $\langle [\frac{1}{1}] , [\frac{1}{2}], [\frac{1}{3}], \dots \rangle$? Suppose $\mathbb{R}$ has the usual topology. Endow $\mathbb{R}$ with an equivalence relation $\sim$ via $x \sim y$ if $x$ and $y$ are irrational or $x \sim y$ if $x = \frac{a}{n}$, $y = \frac{b}{n}$ where $\frac{a}{n}, \frac{b}{n}$ are nonzero, simplest form fractions; finally, we relate $0$ to $\frac{a}{n}$ where $n=1$. That is to say, $\sim$ relates all irrationals together and $\sim$ relates all simplest-form, relatively prime fractions together. A simpler way to describe $\sim$ would be to relate real numbers to each other if they have like outputs via Thomae's function $t$, i.e. $x \sim y \iff t(x) = t(y)$
The quotient space $\mathbb{R} / \sim$ comprises a countable collection of equivalence classes. Denote the class of irrationals $[\star]$ and denote the class of relatively prime denominator-$n$ fractions by $[\frac{1}{n}]$.
$$\mathbb{R} / \sim \hspace{0.2cm} = \Big\{[\star],\Big[\frac{1}{1}\Big], \Big[\frac{1}{2}\Big], \Big[\frac{1}{3}\Big], \dots \Big\}$$
Explicitly, these equivalence classes appear as such
$$\Big[ \frac{1}{1} \Big] = \mathbb{Z}, \hspace{0.2cm} \Big[ \frac{1}{2} \Big] = \Big\{ \dots-\frac{5}{2}, - \frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots \Big\}$$
Recall also that in general

If $X$ is a topological space and $X / \sim$ is a quotient space of $X$, then $U$ is open in $X / \sim \iff \displaystyle\bigcup_{[x] \in U} [x]$ is open in $X$

The question is this: Is $[\star]$ a limit point of the sequence $\big\langle [\frac{1}{n}] \big\rangle_{n=1}^\infty$ in $\mathbb{R} / \sim \hspace{0.2cm}$? To show this, we need to show that for every open nbhd $U$ of $[\star]$, there is $M \in \mathbb{N}$ so that when $n \geq M, [\frac{1}{n}] \in U$.
First, note that $U \neq \{[\star]\}$, i.e. $U$ cannot be a singleton since $[\star] = \mathbb{R} \setminus \mathbb{Q}$ is not open in $\mathbb{R}$. Moreover, $U$ also cannot contain a finite number of $[\frac{1}{n}]$ classes because the union of that finite collection of classes would be neither open nor closed in $\mathbb{R}$. Hence, $U$ must be the entire quotient space with a finite number of  $[\frac{1}{n}]$ classes taken away; we would write this as $U = (\mathbb{R} / \sim) \setminus \{[\frac{1}{n_1}], [\frac{1}{n_2}], [\frac{1}{n_3}], \dots, [\frac{1}{n_k}]\}$. In $\mathbb{R}$, the union of $U$ would take on the appearance of the real line with infinitely many holes taken out of it, the holes of which are at least a finite distance apart from each other, which could be expressed as the union of disjoint, open intervals.
How can I show that $U = (\mathbb{R} / \sim) \setminus \{[\frac{1}{n_1}], [\frac{1}{n_2}], [\frac{1}{n_3}], \dots, [\frac{1}{n_k}]\}$? If this can be shown, then I can show that $[\star]$ is a limit point of $\big\langle [\frac{1}{n}] \big\rangle_{n=1}^\infty$
 A: The complement of $U$ is not necessarily finite. For a counterexample,
$$ U = (\mathbb{R} / \sim) \setminus \bigcup_{n=1}^\infty \left[\frac{1}{n!}\right] $$
The only integers relatively prime to $n!$ are of the form $m(n!) \pm 1$, so the union of elements of $U$ is $\mathbb{R} \setminus V$ where
$$ V = \left\{ m \pm \frac{1}{n!} \mid m \in \mathbb{Z}, n \in \mathbb{N} \right\} $$
$V$ is closed in $\mathbb{R}$, and its limit points are the integers. So $U$ is open in $\mathbb{R} \setminus \sim$.
However, the definition of "a limit point" of a sequence $(a_n)$ doesn't require $a_n \in U$ for all $n \geq M$. That's the definition of "converges to a limit". This example $U$ shows that the sequence $\left(\left[\frac{1}{n}\right]\right)_{n=1}^\infty$ does not converge to $[\star]$. But we call $b$ a limit point of $(a_n)$ if at least one $a_n$ is in every open neighborhood of $b$. Since every non-empty open set in $\mathbb{R}$ contains a rational number, every open set in $\mathbb{R} / \sim$ containing $[\star]$ also contains some $\left[\frac{1}{n}\right]$, so $[\star]$ is a limit point of $\left(\left[\frac{1}{n}\right]\right)_{n=1}^\infty$.
