TITLE: Which sequences converge in a cofinite topology and what is their limit?
QUESTION [15 upvotes]: This is an exercise from an earlier calculus 1 reading at my university:

Let $X$ be a space containing infinitely many elements. In the cofinite topology, a set $\Omega$ is open iff $\Omega = \emptyset$ or $\Omega^c$ only contains finitely many elements. Which sequences converge and what is their limit?

As I had absolutely no idea how to approach this exercise, I looked up the solution and I found it quite frustrating that I didn't even understand the solution. The writer of the solution states that for any sequence $(a_n)$ in $X$, there are three possible cases:

a) There exists no value which the sequence takes infinitely many times.
b) There exists exactly one value which the sequence takes infinitely many times.
c) There exist two or more values which the sequence takes infinitely many times.

In the first case, the writer further states that the sequence converges to any value $a \in X$, in the second case, the sequence only converges to the value taken infinitely many times and in the last case, the sequence diverges. Obviously, the writer left out a proof or even a reasoning which might help a reader to understand why this is correct.
Can anyone help me out by explaining why this holds true? Thank you very much in advance.

REPLY [9 votes]: Say $X$ is a set endowed with the cofinite topology $\mathcal T$. Say $\{x_n\}_{n=1}^\infty$ is a sequence in $X$. We claim that $\{x_n\}$ converges in $(X,\mathcal T)$ if and only if there is at most one value in $\{x_n\}$ which occurs infinitely many times.
For the $(\Longrightarrow)$ direction, suppose that $\{x_n\}$ converges. Suppose $a$ and $b$ are distinct values in $\{x_n\}$ which occur infinitely many times. Observe that $X\setminus\{b\}$ is open and contains $a$. But there are infinitely many $x_n\notin X\setminus\{b\}$ and so $\{x_n\}$ cannot converge to $a$. Analogously, we see that $\{x_n\}$ cannot converge to $b$. But we assumed that $x_n\to c$ for some $c\in X$. Considering $X\setminus\{a,b\}$, we see by an analogous argument to that for $a$ and $b$, that $\{x_n\}$ cannot converge to $c$. Therefore, if $\{x_n\}$ converges, there is at most one infinitely repeating term of the sequence.
For the $(\Longleftarrow)$ direction, assume there is at most one infinitely repeating term of the sequence. There are two cases:

There are no infinitely repeating terms. We claim that $\{x_n\}$ converges to every $x\in X$. Indeed, if $N(x)$ is a neighbhorhood of $x$, then $X\setminus N(x)$ is finite. As there are no infinitely occurring terms, there must be some $m\in\mathbb N$ such that $x_n\in N(x)$ whenever $n>m$.
There is exactly one infinitely repeating term. Call this value $a$. We claim that $\{x_n\}$ converges to $a$ and only to $a$. If $N(a)$ is any neighborhood of $a$, then $X\setminus N(a)$ is finite. As no other term repeats infinitely many times, there is $m\in\mathbb N$ such that $m>n$ implies $x_n\in N(a)$. So $x_n\to a$. Now suppose $x_n\to a'$ for $a\neq a'$. Then $X\setminus\{a\}$ is open and contains $a'$, but there are infinitely many $x_n\notin X\setminus\{a\}$. So $\{x_n\}$ cannot converge to $a'$.