How many "co-" topologies are out there? So far I have learned about $\tau_{co-finite}$ and $\tau_{co-countable}$
Are there any other co-related topologies like...$\tau_{co-infinite}$?
In general, what is the condition we need to have a co-topology?
 A: If $\mathcal{C}$ is any collection of subsets of a set $X$, you could try to define a "co-$\mathcal{C}$" topology on $X$ whose open sets are exactly the complements of elements of $\mathcal{C}$.  This actually is a topology iff $\mathcal{C}$ satisfies the axioms for closed sets (which are just "complemented" versions of the axioms for open sets): $\mathcal{C}$ must be closed under arbitrary intersections and finite unions.
A: Mikhail's answer has the right idea. Here's how to do it rigorously:

Definition. Let $X$ denote a set and $\kappa$ denote a cardinal number. Then $\tau(X,\kappa)$ is the collection of all subsets $A$ of $X$ such that $|A| < \kappa$ or $A = X$.
Proposition. If $\kappa \geq \aleph_0$, then $\tau(X,\kappa)$ satisfies the closed-set axioms for a topology on $X$.

For instance, $\tau(X,\aleph_0)$ is the cofinite topology, and $\tau(X,\aleph_1)$ is the co-countable topology.
As there are infinitely many different cardinal numbers, we can make infinitely many different such spaces. We do "need" (to get something interesting) that $|X| \ge \kappa$, or else $\tau(X,\kappa)$ is just the discrete topology. In that case, for different $\kappa$ and $X$ or $X$ of different size, we get non-homeomorphic spaces, and all these spaces are $T_1$ but not $T_2$. And for all $\kappa \ge \aleph_1$ they have no convergent subsequences except the trivial ones that are eventually constant.
A: You can define $\tau_{co-\sigma}$. Where $\sigma$ is any infinite cardinality. Define open sets as any subset which has complements with cardinality strictly less than $\sigma$. Check that it is indeed a topology!
Your examples are special cases, where $\sigma$ is countable set and continuum set respectively, if you assume continuum hypothesis.
