Is $(\mathbb{R},\tau_{co})$ compact where $\tau_{co}$ is the cocountable topology on $\mathbb{R}$?
I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says:
$\mathbb{N}-\{n\}$ is infinite countable for all $n\in\mathbb{N}$.
Then $V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}$ for all $n\in\mathbb{N}$.Then $U=\{ V_n : n\in\mathbb{N}\}$ is a open covering of $(\mathbb{R},\tau_{co})$.
$U$ does not contain a countable subcover since $V_n\cap\mathbb{N}=\{n\}$ for each $n\in\mathbb{N}$ and $\mathbb{N}$ is infinite.
How would you solve that problem?
Best Answer
For $q \in \mathbb Q$, let $\mathcal O_q = (\mathbb R \setminus \mathbb Q) \cup \{q\}$ i.e. the set of all irrational numbers and $q$. Then, $\mathcal O_q$ is cocountable and $\mathbb R = \bigcup_{q\in \mathbb Q} \mathcal O_q$ but no finite subcover would cover all of the rational numbers.