Suppose that $X$ is a (no finite) set endowed with the cofinite topology. Is this space a sequential space? My definition is that a topological space $X$ is a sequential space if and only if the open sets $U$ are such that for any sequence $(x_n)$ such that $\lim_n x_n\subseteq U$ then there is $n_0\in\mathbb N$ such that $x_n\in U$ whenever $n\geq n_0$.
My doubt is generated by the condition $\lim_n x_n\subseteq U$:
Suppose that the sequence has infinit different terms and let $x\in X$. Any open set $U$ containing $X$ is such that all but a finite number of points of the whole space belong to $U$. In particular, $U$ contain all but a finite number of members of the sequence, i.e. $x$ is a limit point for the sequence. This implies that $\lim_n x_n =X\not\subseteq U$.
Does this mean that I have to restrict to sequences such that $x_n=x_0$ whenever $n$ is greater than some $n_0\in\mathbb N$? (Notice that I need the sequence to have a limit, so it is not enough for the sequence to have contained, as a set, in a finite subset, e.g. $x_n=(-1)^n)$.
If I restrict myself to this kind of sequences then every closed set is sequentially closed, because the sequence is made of elements of the set.
Best Answer
In any space an open set is sequentially open by definition.
The main observation for the cofinite topology (on an infinite set) is the following: there are exactly three types of sequence:
a. $(x_n)$ has infinitely many terms, so $\{x_n: n \in \mathbb{N}\}$ is infinite. Then $\lim (x_n) =X$, i.e. every $p \in X$ is a limit of this sequence.
b. $(x_n)$ has finitely many distinct terms and is eventually constant, i.e. there is some $N$ and some $x \in X$ such that $\forall n \ge N: x_n = x$ and in that case $\lim (x_n) = \{x\}$.
c. $(x_n)$ has finitely many terms and is not eventually constant. Then this sequence has no limit, so $\lim_n (x_n) = \emptyset$.
Now suppose that $U$ is sequentially open, so for any sequence $(x_n)$ in $X$ we have that if $\emptyset \neq \lim_n (x_n) \subseteq U$ then $(x_n)$ is eventually in $U$. Note that the non-emptyness is important (see also wikipedia for another formulation of this definition: for all sequences $(x_n)$ and all $x$, if $x_n \to x$ and $x \in U$ then $(x_n)$ is eventually in $U$.)
We want to show that $U$ is open. If $U$ is empty, we are done, so assume $U\neq \emptyset$. Suppose we have a sequence with $\lim_n (x_n) \subseteq U$. If it is of type a. then $\lim (x_n) = X$ and $X = U$ and we are done. If it is of type b., we are also done as in particular $x \in U$ and $(x_n)$ is eventually $x$. Type c. is not an allowed "test sequence" for sequential openness. So in all cases $U$ is open and we're done.
We could also test (more easily) sequentially closed sets: suppose $F$ is infinite and sequentially closed. If $F \neq X$ pick $p \notin F$ and countably many different $x_n \in F$ for all $n$. Then $(x_n)$ would be a sequence in $F$ that converges to $p$ while $;p \notin F$, so $F$ would be not be sequentially closed at all. This contradiction came from assuming $F \neq X$, so $F=X$. Hence all sequentially closed sets are finite or $X$, as required.
So indeed the cofinite topology is a sequential space.