# 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.

-
You will have to give the definitions of open/closed to make this question answerable from the point of view of the class. There are texts that say $G$ is open if every point of $G$ is contained in a ball contained in $G$, and that $F$ is closed if R^n\F is open (by def). There are texts that say that $F$ is closed if it contains all of its limit points, and that $G$ is open if R^n\G is closed (by def). There are texts that say $G$ is open if every point is in a ball in $G$, $F$ is closed if it contains its limit points. In the latter case it is a thm that open iff complement closed. – Jonas Meyer Mar 31 '11 at 1:44
Wow, I never knew there was that degree of variation among textbooks. The definition given to us is that a closed set is one that contains all of it accumulation points. – user8921 Mar 31 '11 at 1:57
Thanks. Now, how about the definition of open? (By the way, I am just being cautious. I guessed that closed probably meant for you containing all accumulation points and open probably meant every point is contained in a ball in the set, but there are numerous possible characterizations, so when proving such a basic fact one should be clear on the definition.) – Jonas Meyer Mar 31 '11 at 1:58
Oh, right, guess I should have included that too! 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. Let me know if you need any more clarification. I'm just a bit in over my head with all this :) – user8921 Mar 31 '11 at 2:11

The following are equivalent:

1. $G$ is open.

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

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

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

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

6. $\mathbb{R}^n\setminus G$ 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 $n+1$.

-

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.

-
I really appreciate the help, gentlemen. I think I'm beginning to understand the steps by which I can tackle this problem. Helps with some other concepts too. Thanks again! – user8921 Mar 31 '11 at 3:20