The Countable Complement Topology

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)
\begin{align} \quad \left ( \bigcup_{i \in I} U_i \right )^c = \bigcap_{i \in I} U_i^c \end{align}

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)
\begin{align} \quad \left ( \bigcap_{i=1}^{n} U_i \right )^c = \bigcup_{i=1}^{n} U_i^c \end{align}

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.