What are the basis for co-countable and co-finite topologies I have just learned co-countable and co-finite topologies
But it was never mentioned what are the bases for these topologies and I cannot seem to find any reference to it anywhere.
For example, in the standard topology $\tau_{standard}$ on $\Bbb{R}$, a basis is $\mathcal{B} = \{(a,b) \subset \Bbb{R} | a, b \in \Bbb{R}\}$
Is there something similar for those topologies?
 A: A basis is just one tool for describing a topology.  Most topologies have many different bases that generate them, but most of these bases aren't useful.  The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis.  After all, what could be easier to think about than simply saying a set has finite (or countable) complement?  The open sets themselves already have a very simple structure.  Moreover, it's not as though open sets are built out of small "balls" around each point.  For instance, in the cofinite topology, if $U$ is open and $x\in U$ and you take an open set $V$ such that $x\in V\subseteq U$, then $V$ must already contain all but finitely many points of $U$.  So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $\mathcal{B}=\{U\subset X:|X\setminus U|=1\}$ of complements of single points.  Every cofinite set is a finite intersection of these sets, so they do form a subbasis.  In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $\mathcal{B}$, or a map $f:Y\to X$ is continuous iff $f^{-1}(U)$ is open for each $U\in\mathcal{B}$.
