# Cocountable Topology

Given a space $X$, there are many ways to define a topology. More common methods are one of the following:

• explicitly list out all the open sets
• define the rules for a subset $U \subseteq X$ to be open
• define the basis/subbasis of the topology (Note: The basis here refers to topological basis, this is different than a vector space basis that we have learnt in linear algebra.)

More often than not, we will work with topological spaces that has infinitely many open sets. In this case it will be hard to explicitly list out all the open sets in the topological space. In this post, we will look at a topology call cocountable topology, and luckily, we can describe the open set of this topology, thus we do not need to describe a topological basis.

Given a ground set $X$, the cocountable topology $\tau$ on $X$ is the empty set together with the collection of subsets $U \subseteq X$ such that $X \setminus U$ is countable, namely the complement of $U$ is countable. For example, if $X = \{1,2,3\}$, then under the cocountable topology, none of the subsets of $X$ is open.

Now let’s assume that $X$ is an uncountable set.  By definition, $\emptyset, X \in \tau$. Then, let $\{U_{\lambda}\}_{\lambda \in \Lambda} \subseteq \tau$ be a collection of open sets in $X$ for some index set $\Lambda$. Thus $X \setminus U_{\lambda}$ is countable for all $\lambda \in \Lambda$. Now consider

$X \setminus \bigcup_{\lambda \in \Lambda} U_{\lambda} = \bigcap_{\lambda \in \Lambda} [X \setminus U_{\lambda}] \subseteq X \setminus U_{\lambda_0}$

by De Morgan’s law, where $\lambda_0 \in \Lambda$ is arbitrary. Since $X \setminus U_{\lambda}$ is countable for all $\lambda \in \Lambda$ and intersections will only result in smaller set, therefore $X \setminus \bigcup_{\lambda \in \Lambda} U_{\lambda}$ must be countable. Thus we may conclude that $\bigcup_{\lambda \in \Lambda} U_{\lambda} \in \tau$.

Finally, let $\{U_i\}_{i=1}^{n} \subseteq X$ be a finite collection. Then by De Morgan’s law, we have $X \setminus \bigcap_{i=1}^{n} U_i = \bigcup_{i=1}^{n} (X \setminus U_i)$. Notice that finite union of countable sets is still countable (in fact, countable union of countable sets is countable), thus $X \setminus \bigcap_{i=1}^{n} U_i$ is countable and $\bigcap_{i=1}^{n} U_i \in \tau$.

Therefore $\tau$ is indeed a topology.