A subset $G$ of $\mathbb{R}^n$ is open iff the complement of $G$ is closed So I'm covering material for my upcoming final exam, and I have a sneaking suspicion that my teacher will ask us to prove the following theorem: 

A subset $G$ of $\mathbb{R}^n$ is open iff the complement of $G$ is closed. 

He's hinted at it a couple times, and honestly, I don't know where to start. Thanks for the help.
Definitions (copied from comments): A set is open if every point of the set is an interior point, meaning that the set contains some ball of positive radius at any one of the interior points.  A closed set is one that contains all of its accumulation points. 
 A: The following are equivalent:


*

*$G$ is open.

*For all $x\in G$, there is an open ball $B$ such that $x\in B\subseteq G$.

*For all $x\in G$, there is an open ball $B$ such that $x\in B$ and $B\cap(\mathbb{R}^n\setminus G)=\emptyset$.

*For all $x\in G$, $x$ is not an accumulation point of $\mathbb{R}^n\setminus G$.

*For all $x\in \mathbb{R}^n$, if $x$ is an accumulation point of $\mathbb{R}^n\setminus G$, then $x$ is in $\mathbb{R}^n\setminus G$.

*$\mathbb{R}^n\setminus G$ is closed.
The equivalence of 1&2 and of 5&6 is by definition.  For the rest, see if you can find a straightforward proof that statement $n$ is equivalent to statement $n+1$.
A: Assume $U\subseteq \mathbb{R}^n$ is open and let $p\in U^c$ be an accumulation point.  We need to show $p\in U^c$.  If $p\notin U^c$, then $p\in U$ so there is an open set $V$ such that $p\in V\subseteq U$.  But then $p\in V$ and $V\cap U^c = \emptyset$ which contradicts the fact that $p$ is an accumulation point of $U^c$.
Conversely, assume $U^c$ is closed and let $p\in U$.  We want to find an open neighborhood $V\subseteq U$ with $p\in V$.  Assume no such $V$ exists which means for every $V$, we have $V\cap U^c \neq \emptyset$.  But this says exactly that $p$ is an accumulation point of $U^c$.  Since $U^c$ is closed, we conclude $p\in U^c$, which is a contradiction.  Hence, there is some open set $V$ for which $p\in V\subseteq U$.  Since $p$ is arbitrary, we conclude $U$ is open.
