Convergence of sequences in topological spaces Consider R with co-countable topology then for Q'(set of irrational numbers) subset of R ,how can we show that there is no sequence in Q' converges to 0?
If $(x_n)$ is a sequence of irrational numbers converging to $0$ then $\mathbb R \setminus \{x_1,x_2,..\}$ is an open set in this topology containing $0$ so it must contain $x_n$ for $n$ sufficiently large. But it contains no $x_n$!
Any sequence $x_n$ converges to $0$ if for all open $\theta$ with $0\in \theta$ then exist $N$ a natural number with $x_n \in \theta$ for all $n\geq N$.

Let $\theta = \mathbb{R}\setminus \{x_n\}$. Then $\theta$ is an open set, because his complement is a contable set, it's mean, $\{x_n\}$ is contable. For the other hand, for all $n\in\mathbb{N}$, $x_n \not\in\theta$.

A more general result :

If $x$ is a rational number, there isn't any sequence $x_n$ of irrational numbers with $x_n\to a$.