Does the sequence $\langle \frac{1}{n} \rangle$ converge to $0$ under the discrete rational extension? 
Let $\mathfrak{T}$ be the usual topology on $\mathbb{R}$. The discrete rational extension is a topology on $\mathbb{R}$ given by $$\mathfrak{A}=\{U \cup Q: U \in \mathfrak{T}, Q \subset \mathbb{Q}\}$$

Suppose $\mathbb{R}$ is endowed with the discrete rational extension topology, $\mathfrak{A}$. Then, does the sequence $\frac{1}{n} \to 0$, $n \in \mathbb{N}$? Under this topology, we now have that $\mathbb{Q}$ as a subspace of $\mathbb{R}$ is discrete, hence $\{q\}$ is open for every $q \in \mathbb{Q}$. If $\frac{1}{n} \to 0$, then every open nbhd of $0$ would contain a tail end of the sequence.
However, as said, $\{0\}$ is open in $\mathbb{R}$. Then $\frac{1}{n} \not\to 0$. Moreover, I suspect no sequence can converge to a rational number. Maybe I will generalize my question.

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*Can sequences converge to rational numbers under this topology?


I believe the following is true here:
Claim. Let $q \in \mathbb{Q}$, then a sequence $(x_n)_{n \in \mathbb{N}}$ converges to $q$ if and only if it is eventually constant.
Proof. Clearly, if the sequence is eventually constant it converges to $q$. Conversely, if it is not eventually constant then for every $N \in \mathbb{N}$ we find $k > N$ such that $x_k \neq q$. Since singleton sets of rationals are open in this topology, we thus have that $\{q\}$ is an open neighborhood of $q$ that does not contain any tail of the sequence $(x_n)_{n \in \mathbb{N}}$, therefore $(x_n)_{n \in \mathbb{N}}$ does not converge to $q$.
