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

general-topologymetric-spaces

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.

Best Answer

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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