Let $X$ be a nonempty set. If the cocountable topology $\tau$ is considered on $X$, then I want to find all compact and sequential compact subsets of $X$.
Definitely only finite subsets of $X$ are compact. But I am in confusion to find all sequentially compact sets. Is there any infinite set which is sequentially compact? Plese help.
Again, the only sequentially compact sets are finite. In fact, you can show that the only convergent sequences are those that are eventually constant :
If $(x_n) \subset X$ such that $x_n \to l$, then set $A = \{x_n : x_n \neq l\}$, then $A$ is countable, and $U = X\setminus A$ is an open neighbourhood of $l$. So $\exists N \in \mathbb{N}$ such that $x_n \in U$ for all $n\geq N$. But that implies that $x_n \notin A$, which means $x_n = l$ for all $n\geq N$
Now, let $Y \subset X$ be an infinite set, so choose a sequence $(x_n) \subset Y$ with distinct terms. Every possible subsequence of $(x_n)$ has distinct terms, and so does not converge. So, $Y$ is not sequentially compact.
