Exercise 2.1.2 (Cofinite topology)

Proof. Let $T$ be as defined in the exercise, i.e.,

We verify that $T$ is indeed a topology.

1.

$\varnothing \in T$ by construction and $X\in T$ since its complement $X\setminus X=\varnothing$ is trivially finite.

2.

Let ${(A_j)}_{j\in J}$ be a family of sets in $T$. To see whether ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j\in T$, consider the cardinality of its complement. By de Morgan’s laws and by cardinal arithmetic, $$\begin{array}{llll}\hfill \#\left(X\setminus \bigcup _{j\in J}{A}_j\right)& =\#\left(\bigcap _{j\in J}X\setminus A_j\right)& \hfill & \\ \hfill & \le \#\left(X\setminus A_1\right).& \hfill & \end{array}$$

Since $A_1\in T$, its complement must be finite, and so is the complement of ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$.

3.

Let ${(A_n)}_{n=1}^N$ be a finite family of sets in $T$. We look at the complement of their intersection. Using de Morgan’s laws and basic cardinal arithmetic again, we obtain $$\begin{array}{llll}\hfill \#\left(X\setminus \bigcap _{1\le n\le N}{A}_n\right)& =\#\left(\bigcap _{1\le n\le N}X\setminus A_n\right)& \hfill & \\ \hfill & \le \sum _{1\le n\le N}\#\left(X\setminus A_n\right)<\infty .& \hfill & \end{array}$$

In other words, ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{1\le n\le N}{A}_n$ is contained in $T$.