I found this question online (not homework) and wanted to check my answer.
Let $$A_1 = \{r: r \textit{ rational, and } \lim \text{inf } x_n <r < \lim \text{sup } x_n \}$$ $$A_2 = \{r: r \textit{ rational, and } \lim \text{inf } x_n >r\}$$ $$A_3 = \{r: r \textit{ rational, and } \lim \text{sup } x_n <r\}$$ State a simple condition involving the three sets above, that is equivalent to "$(x_n)$ converges to a finite value".
I believe that the answer is $A_1 \cup A_2 \cup A_3 = A_2 \cup A_3$, or simply that $A_1=\emptyset$.
My argument is that if $(x_n)\to p$, then $\lim \text{sup }=\lim \text{inf } = p$. Therefore set $A_1$ is empty, $A_2 = \{q\in\mathbb{Q}:q<p\}$ and $A_3 = \{q\in\mathbb{Q}:q>p\}$
Meanwhile, if $(x_n)$ does not converge, $A_1$ is not empty since $\lim \text{inf } < \lim \text{sup }$ and $\mathbb{Q}$ is dense in $\mathbb{R}$.
Is my reasoning generally correct? Thank you very much.