Defining convergence in terms of the set of rationals in the gap between $\liminf$ and $\limsup$ Consider a real sequence $x_n$ and the set $A = \{ r | \liminf x_n < r < \limsup x_n \, , \,  r \in \mathbb Q\}$. Is it true that $|A| = 0$ if and only if $x_n$ converges to a real limit?
My attempt:
Note that if $\liminf x_n < r < \limsup x_n$ for some real $r$ then there exists a $q \in \mathbb Q$ close enough to $r$ such that $\liminf x_n < q < \limsup x_n$ since the rationals are dense in the reals. Also, if there exists a rational $q$ such that $\liminf x_n < q < \limsup x_n$ then the inequality is satisfied for a real value since $\mathbb Q \subset \mathbb R$.
So, the set $A$ above has cardinality $0$ if and only if the set $B$ has cardinality $0$ where $B = \{r | \liminf x_n < r < \limsup x_n, r \in \mathbb R\}$.
But $B$ has cardinality $0$ only when $\liminf x_n = \limsup x_n$ which implies that $x_n$ converges to a real limit.
I could use some feedback on my reasoning.

If $|A|$ means measure of $A$ then $|A|=0 $ for any sequence $x_n .$
If $|A|$ means the cardinality of $A$ then $|A|=0$ if and only if $x_n$ converges to finite or infinite limit.
