Closed subspace of $\mathbb{R}\times\mathbb{Q}$ In a proof, it says that the subspace $\{(m+\frac{1}{2n},\frac{r_n}{m}) | m,n\in \mathbb{N} \}$ of $\mathbb{R}\times \mathbb{Q}$ is closed, where $(r_n)_{n\in \mathbb{N}}$ is a strictly decreasing sequence in $\mathbb{Q}$ converging to $\sqrt{2}\in\mathbb{R}$. How can I see this?
 A: Denote $V = \mathbb R \times \mathbb Q$. We have
$$S=\{(m+\frac{1}{2n},\frac{r_n}{m}) | m,n\in \mathbb{N} \} = \bigcup_{m \in \mathbb N} S_m$$
where $$S_m=\{(m+\frac{1}{2n},\frac{r_n}{m}) | n\in \mathbb{N} \}.$$
Each of the $S_m$ is a set of discrete points without any limit point in $\mathbb R \times \mathbb Q$. Hence is closed.
Also for all $m \in \mathbb N$, you have $S_m \subseteq [m, m+1/2] \times \mathbb Q = T_m$. As the $T_m$ are closed and their mutual distances are at least $1/2$, a limit point of $S$ has to be a limit point of one of the $S_m$.
As we proved above that the $S_m$ have no limit points in $V$, we can conclude that $S$ doesn't have any limit points in $V$ and hence is closed.
A: One possible way to show this is to see $\mathbb{R\times Q}$ as a subspace of $\mathbb{R\times R}$. You can see that the set
$$A:=\left\{\left(m+\frac{1}{2n},\frac{r_n}{m}\right) : m,n\in\mathbb{N}\right\}\cup \{(m,\sqrt{2}/m):m\in\mathbb{N}\},$$
which is a topological closure of the given set over $\mathbb{R\times R}$, is closed.
Moreover, our set is equal to $A\cap \mathbb{R\times Q}$. Could you conclude why the given set is closed in $\mathbb{R\times Q}$?
