To prove that a given set is a topology

Question

Problem

Let $\mathbb{N}$ be the set of all positive integers and let $\tau$ consists of $\emptyset$ and each subset $S $ of $\mathbb{N}$ such that complement of $S$ in $\mathbb{N}$ , $\mathbb{N}$\ $S$, is a finite set. Prove that ($\mathbb{N}$ ,$\tau$) is a topological space

Attempt

1. $\mathbb{N} ,\emptyset \in \tau$

2. $\mathbb{N}- \cup_{\alpha} S_\alpha$= $\mathbb{N} \cap (\cup_{\alpha} S_\alpha)^c =\cap_{\alpha}(\mathbb{N}-S_\alpha)$

3. $\mathbb{N} - \cap_{i=1}^{n}S_i= \cup_{i=1}^{n}(\mathbb{N}-S_i)$, which is finite. Is this correct?

Answer 1

If we have the union of open sets $S_\alpha$, then either they're all empty (boring special case, that we can skip, as then the union is empty and there's nothing to prove, just as the union over an empty index set is also empty) or at least one $S_\alpha$, say $S_{\alpha_0}$, has a finite complement.

Then:

$$\mathbb{N} - (\bigcup_\alpha S_\alpha) =\bigcap_\alpha (\mathbb{N}-S_\alpha) \subseteq \mathbb{N} - S_{\alpha_0}$$

first by de Morgan then by noting the intersection is a subset of every one of the intersecting sets, including $\mathbb{N}-S_{\alpha_0}$.

A subset of a finite set is finite so the union has finite complement and thus is open.

Answer 2

Since each $S_i\in\tau$ for all $i=1,2,...,n$ so each of $\mathbb{N}-S_i$ is a finite set $\forall i$. and since finite intersection of finite set is finite (if it is null set then vacuously finite) hence $\displaystyle\bigcap_{i=1}^n\mathbb{N}-S_i$ is finite.

Note: It is known as Cofinite topology