How can I prove $\tau$ is a topology on $\mathbb{N}$

Question

Consider the set $X = \mathbb{N}$, and let $\tau$ be the collection of all subsets $A \subset \mathbb{N}$ for which $\mathbb{N}\setminus A$ is finite, along with the empty set. I want to show $\tau$ is a topology on $\mathbb{N}$.

I know that the requirements for a set to be a topology is:

1. $\emptyset$ and $X$ are in $\tau$

2. The union of the elements of any subcollection of $\tau$ is in $\tau$

3. The intersection of the elements of any finite subcollection of $\tau$ is in $\tau$.

For this problem, I know that the first condition holds.

Can someone please help me show that the other two conditions for a topology hold as well so that I can prove that $\tau$ is a topology on $\mathbb{N}$

Answer 1

Let $\tau$ be a family of subsets of $\mathbb N$ as follows: $$\tau=\{A: \mathbb N \setminus A \text{ is finite or } A=\emptyset \}$$ We want to show that $\tau$ is a topology on $\mathbb N$.

1. First, we show that both $\emptyset$ and $\mathbb N$ are in $\tau$. By definiton, $\emptyset \in \tau$. Since $\mathbb N \setminus \mathbb N = \emptyset$, $\mathbb N$ is also in $\tau$, so this condition is satisfied.
2. Second, we show that any arbitrary union of elements in $\tau$ is also in $\tau$. Let $\{A_\alpha\}$ be an arbitrary family of sets in $\tau$. If $A_\alpha = \emptyset$ for all $\alpha$, then $\bigcup A_\alpha = \emptyset$ and therefore is a member of $\tau$. Now consider the case where at least one $A_\alpha \ne \emptyset$. We can denote this nonempty set as $A_{\alpha_1}$. Because $A_{\alpha_1} \in \tau$, we know that $\mathbb N \setminus A_{\alpha_1}$ is finite, and since $A_{\alpha_1}\subset \bigcup A_\alpha$, we see that $\mathbb N \setminus \bigcup A_\alpha$ must be finite as well and is therefore a member of $\tau$. This satisfies our second condition.
3. Finally, we show that the intersection of finitely many elements of $\tau$ is also in $\tau$. Let $(A_1, A_2,...,A_n)$ be a finite collection of sets in $\tau$. If any $A_i=\emptyset$, then $\bigcap_{1 \le i \le n}A_i = \emptyset$ and is therefore in $\tau$. Now consider the case where all $A_i$ are nonempty sets. Since $A_i \in \tau$, we know that $\mathbb N \setminus A_i$ is a finite set. Since $\bigcap_{1 \le i \le n}A_i = \mathbb N \setminus \left( \bigcup_{1 \le i \le n} \mathbb N \setminus A_i \right)$, and since a finite union of finite sets is finite, we see that $\bigcap_{1 \le i \le n}A_i$ has a finite complement and therefore must be in $\tau$, satisfying condition 3.

This proves that $\tau$ is a topology on $\mathbb N$. This particular topology is known as the cofinite topology and can be defined on any nonempty set, not just $\mathbb N$.