Let $\tau = \{\varnothing\} \cup \{G \subseteq \mathbb R: \mathbb R-G \ $is finite $\} $ and $(\tau,\mathbb R)$ is the topology of the cofinite space. Find a compact set $K \subseteq \mathbb R $ that is not closed.
We have just started topology and I am a bit stumped.
My ideas: As I need to find a set $K \subseteq\mathbb R $ whereby $K^{C}$ is finite. I would naturally suggest $\mathbb R-\{0\}$ or excluding some other finite subset, but then I get trouble in proving that it is compact. Any hints?
Take any countable subset $C$ of $\mathbb R$. Since it is neither $\mathbb R$ nor a finite set, it is not closed.
However, it is compact. Let $(A_\lambda)_{\lambda\in\Lambda}$ is an open cover of $C$. Take $c\in C$. Then there is a $\lambda_0\in\Lambda$ such that $c\in A_{\lambda_0}$. In particular, $A_{\lambda_0}\neq\emptyset$. SInce it is an open set, $\mathbb{R}\setminus A_{\lambda_0}$ is finite. So, only finitely many elements $c_1,\ldots,c_n$ of $C$ don't belong to $a_{\lambda_0}$. For each $k\in\{1,2,\ldots,n\}$, take $\lambda_k\in\Lambda$ such that $c_k\in A_{\lambda_k}$. Then $\bigl\{A_{\lambda_k}\,|\,k\in\{0,1,\ldots,n\}\bigr\}$ is a finite subcover of $(A_\lambda)_{\lambda\in\Lambda}$.