Proof verification that the sequence $x_n = \frac{1}{n}$ converges to every point of $\mathbb{R}$ on the cofinite topology

Question

Let $a \in \mathbb{R}$. Then any open set $U$ in the cofinite topology containing $a$ is of the form $U = \mathbb{R} - \{\alpha_1, \alpha_2, \cdots, \alpha_p\}$ for some $p \in \mathbb{N}$ (and of course $\alpha_i \neq a \ \forall 1 \leq i \leq p$). Now, take $N = c \left(\displaystyle{\frac{1}{\min \alpha_i}}\right) + 1$, where $c(x)$ stands for the ceiling of $x$ (i.e, $c(2.1) = 3$). Then it's clear that $x_n \in U \ \forall n \geq N$ and we're done.

Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $x_n$ is eventually contained in $U$ since $\mathbb{R} - U$ has only finitely many elements and $\{x_n\}_{n \in \mathbb{N}}$ is infinite, but I'm not sure if that's fine too.

Answer 1

Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.

Answer 2

The only thing we are using of the sequence here is that $x_n \neq x_m$ whenever $n \neq m$, so that $\{x_n: n \in \mathbb{N}\}$ is infinite.

If then $U=\mathbb{R}-F$ (so $F \subseteq \mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that $\{x_n: n \ge N\}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F \cap \{x_n: n \in \mathbb{N}\}$, if there are any. And then for all $n \ge N$, $x_n \in U$. This suffices to show that $x_n \to x$ for any $x \in \mathbb{R}$.

So also $1,2,3,4,\ldots \to \pi$, e.g. in the cofinite topology.