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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.

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    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 :)2011-03-31

2 Answers 2

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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 $n+1$.

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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.

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    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!2011-03-31