Which sequences converge in a cofinite topology and what is their limit?

Solution 1:

Case a: Let $(a_n)$ be a sequence in $X$ such that there exists no value which the sequence $(a_n)$ takes infinitely many times. Let $x \in X$ and let $U$ be an open set around $x$. Then $U^c$ is finite. Since $(a_n)$ does not take any value infinitely many times, after some index N, $(a_n) \in U$ for all $n \geq N$. Therefore $(a_n)$ converges to any $x\in X$. (This also proves that $X$ is not Hausdorff.)

Case b: Let $(b_n)$ be a sequence in $X$ such that there exists exactly one value $b \in X$ which the sequence $(b_n)$ takes infinitely many times. Let $U$ be an open set around $b$. Then $U^c$ is finite. Since $(b_n)$ takes at most finitely many values in $U^c$, $(b_n) \rightarrow b$. On the other hand, if $c \neq b$, $X \setminus \{b\}$ is an open set around $c$ that misses infinitely many $b_n$, so that the sequence cannot converge to $c$.

Case c: Let $(c_n)$ be a sequence in $X$ such that there exist two values $c_1$ and $c_2$ which the sequence $(c_n)$ takes infinitely many times. Let $U_1$ and $U_2$ be open sets around $c_1$ and $c_2$, respectively. Then $U_1^c$ and $U_2^c$ are finite by definition. Assume to the contrary, $c_n \rightarrow c$ for some $c \in X$. Let $V$ be an open neighborhood around $c$. Then there exists $N \in \mathbb{N}$ such that $(c_n) \in V$ for all $n \geq N.$ But this contradicts that $c_1$ and $c_2$ occurs in the sequence infinitely many times.

Solution 2:

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:

1. 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 $n>m$.
2. 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 $a'$.

Solution 3:

For case a), we want to show that for any given $a$, and every open set containing $a$, that all the $a_n$ are in the open set as long as $n$ is greater than some $M$ which we get to choose. Since the open set in only missing a finite number of elements, we can put them in order $\{b_1, b_2, \ldots, b_p\}$. Then in the series$ a_n$, since $b_1$ does not appear infinitely many times, there is a last occurrence, call it $b_{1f}$. Similarly there is a last occurrence of $b_2$, call it $b_{2f}$. Then if $n \gt M=\max(b_{1f}, b_{2f},\ldots b_{pf})$, $a_n$ will be in our open set. So the sequence converges to $a$

The arguments for b) and c) are similar.