I am looking at $\mathbb{R}^n$ with the finite complement topology and need to show it's connected.
I know that a connected doesn't have any non-trivial clopen sets.
For $U \in T$ where T is the topology I done $\mathbb{R}^n$ complement the union of all the $U$'s to get the intersection of $\mathbb{R}^n$ complement all the U's. We know this is closed because the U's are open. Then, I did the same vice versa and got that the union of $\mathbb{R}^n$ complement all the U's as closed. Therefore I found no nontrivial clopen subsets. I was wondering if I have done this right and if not how can my answer or method be improved?
Best Answer
By contradiction: Notice that $\Bbb R^n$ is infinite. If it wasn't connected, you get $\Omega_1,\Omega_2$ non-empty, open, and disjoint sets such that $\Bbb R^n = \Omega_1 \cup \Omega_2$. However: $$\Omega_1 \cap \Omega_2 = \varnothing \implies \Omega_1 \subset \Bbb R^n \setminus \Omega_2, \quad \text{and}\quad \Omega_{2}\subset \Bbb R^n \setminus \Omega_1.$$ Since $\Omega_1,\Omega_2 \neq \varnothing$ and we're dealing with the cofinite topology, then $\Bbb R^n \setminus \Omega_1$ and $\Bbb R^n \setminus \Omega_2$ are both finite. By the above, $\Omega_1$ and $\Omega_2$ are finite themselves. Hence $\Bbb R^n = \Omega_1 \cup \Omega_2$ is finite, a contradiction.
The above argument holds for any space $(X, \tau)$ with the cofinite topology (countable complement topology), given that $X$ is infinite (uncountable).