No Proper Subset of the Finite Complement Topology on R is simultaneously open and closed I wrote up a proof that no proper subset of the finite complement on $\mathbb{R}$ is simultaneously open and closed. If someone could critic or give a proof that I could learn from, that'd be great.
Claim. No proper subset of $\mathbb{R}$$_{FC}$ is simultaneously open and closed.
Proof. Let V $\subset$(proper subset of) $\mathbb{R}$$_{FC}$ that is FC-open. We want to prove that $\mathbb{R}$$\setminus$V is not FC-open, so let's consider $\mathbb{R}$$\setminus$V. For $\mathbb{R}$\V to be FC-open means $\mathbb{R}$$\setminus${$\mathbb{R}$\V} must be finite or $\emptyset$. However, $\mathbb{R}$$\setminus${$\mathbb{R}$\V} = V and V is not finite. Hence, $\mathbb{R}$\V $\not$$\in$ FC and therefore not FC-open. Now let B $\subset$ $\mathbb{R}$$_{FC}$ be FC-closed. Then that implies $\mathbb{R}$$\setminus$B is FC-open. Then since $\mathbb{R}$$\setminus$B is FC-open, $\mathbb{R}$$\setminus${$\mathbb{R}$$\setminus$B} must be finite or $\emptyset$. Then, since $\mathbb{R}${$\mathbb{R}$$\setminus$B} = B and B is finite, for B to be FC-open, then $\mathbb{R}$$\setminus$B must be finite. But, $\mathbb{R}$$\setminus$B $\in$ FC implies it is not finite. Hence B is not FC-open.
 A: Your proof is ok, except you go on too long. In the first couple of sentences, you assume $V$ is a proper subset of $\mathbb R$ that is open, and then prove it is not closed. This means $V$ cannot be simultaneously open and closed. You are done! 
There are a couple minor details as well. First, as Daniel Schepler comments, the empty set is open and closed by default (and is the only non-cofinite set in the topology), so you also need to assume $V$ is nonempty. And also, you say $V$ is not finite during the course of your proof, and although that might be somewhat obvious, it would be good to substantiate that. (Also doing so in more detail might have led you to the fact that you needed to assume $V$ was nonempty.)
A: The definition of the finite-complement topology is essentially: a set $F \subseteq X$ is closed iff $F$ is finite or $F=X$.
So if $C\subseteq X$ is closed-and-open and $C \subsetneq X$ (a proper subset) so $C$ must be finite, as it closed and not equal to $X$. Now, $X\setminus C$ is also closed (as $C$ is open) and as it is not finite (as $X$ is infinite and $C$ is finite) we must conclude $X\setminus C = X$, or $C = \emptyset$.
So the only proper clopen subset of $X$ is $\emptyset$. Which suffices to show connectedness.
