The Countable Complement Topology
Recall from the Topological Spaces page that a set $X$ an a collection $\tau$ of subsets of $X$ together denoted $(X, \tau)$ is called a topological space if:
- $\emptyset \in \tau$ and $X \in \tau$, i.e., the empty set and the whole set are contained in $\tau$.
- If $U_i \in \tau$ for all $i \in I$ where $I$ is some index set then $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$, i.e., for any arbitrary collection of subsets from $\tau$, their union is contained in $\tau$.
- If $U_1, U_2, ..., U_n \in \tau$ then $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$, i.e., for any finite collection of subsets from $\tau$, their intersection is contained in $\tau$.
We will now look at an topology that is similar to The Cofinite Topology.
| Definition: If $X$ be a nonempty set. The Countable Complement Topology of $X$ is the collection of subsets $\tau = \{ \emptyset \} \cup \{ U \subseteq X : U^c \: \mathrm{is \: countable} \}$. |
Let's verify that $\tau$ is a topology.
For the first condition, clearly $\emptyset \in \tau$. Furthermore, we note that $X^c = X \setminus X = \emptyset$ which is a countable set, so $X \in \tau$.
For the second condition, let $\{ U_i \}_{i \in I}$ be any arbitrary collection of subsets of $X$ from $\tau$. Then $U_i^c$ is countable for each $i \in I$. By De Morgan's Laws we have that:
(1)The intersection of a collection of countable sets is countable, so $\displaystyle{\left ( \bigcup_{i \in I} U_i \right )^c}$ is countable and so $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$.
For the third condition, let $U_1, U_2, ..., U_n$ be a finite collection of subsets of $X$ in $\tau$. Once again, $U_i^c$ is countable for each $i \in \{ 1, 2, ..., n \}$. By De Morgan's Laws we have that:
(2)The finite union of a collection of countable sets is countable, so $\displaystyle{\left ( \bigcap_{i=1}^{n} U_i \right )^c}$ is countable and so $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$.
Therefore $(X, \tau)$ is a topological space.