Let $\tau=\{\mathbb{Q,I,R},\emptyset\}$. Discuss the convergence or divergence of $\{\frac{1}{n}\}_n$ in this topology.
Before anyone flags this question as a duplicate, I do realize that there is an exact question posted 3 years ago. However I'm not looking for answer so much as an explanation. Or perhaps better stated, a clarification. I believe $\{\frac{1}{n}\}$ converges to $p,\forall p\in\mathbb{Q}$. I believe this is true for $\mathbb{R}$ as well but not for $\mathbb{I}$. Am I correct?
I'll use $\mathbb{P}$ for the irrational numbers, as is common in topology.
So $\tau = \{\emptyset, \mathbb{P},\mathbb{Q}, \mathbb{R}\}$, and it's easy to check this is a topology on $\mathbb{R}$.
Consider the sequence $a_n = \frac{1}{n}$. To check convergence we recall the definition: $a_n \to x$ iff for all open sets $O$ such that $x \in O$, $O$ contains all but finitely many points $a_n$.
The key observation is that if $x \in \mathbb{Q}$ any open $O$ that contains $x$ is either $\mathbb{Q}$ or $\mathbb{R}$, and in either case $O$ contains all $a_n$. So certainly $a_n \to x$. We see that limits of sequences need not be unique. Note also that a constant sequence with rational value also converges to all rationals.
Also, if $x \in \mathbb{P}$, then $\mathbb{P}$ is itself an open set containing $x$ that contains no points $a_n$. So $a_n \not\to x$.
As a bonus if $b_n = \frac{\pi}{n}$, so all $b_n$ are irrational, then $(b_n)$ converges to all $x \in \mathbb{P}$ and does not converge to any $x \in \mathbb{Q}$, in a similar way.
Also, if $c_n$ is defined as $c_{2k} = \frac{1}{2k}, c_{2k+1} = \frac{\pi}{2k+1}$, then $c_n$ converges to no point: no rational point can be a limit as $\mathbb{Q}$ is an open neighbourhood that misses infinitely many $c_n$, and no irrational point can be a limit as the same holds for $\mathbb{P}$.
Reasoning this way we can easily determine the convergence of any sequence. And if any rational is a limit, then all rationals are. Likewise for irrationals.