# Set is open iff complement is closed in $\mathbb{R^n}$

Is my proof of the following correct?

1. A set $A\subseteq \mathbb{R^n}$ is open iff $\mathbb{R^n}-A$ is closed.
2. A set $A\subseteq \mathbb{R^n}$ is closed iff $\mathbb{R^n}-A$ is open.

Proof. Suppose that $A$ is open. We must show that for every $x\notin \mathbb{R^n}-A$ there is a nbhd $N$ with $N\cap (\mathbb{R^n}-A)=\varnothing$. Let $x\notin \mathbb{R^n}-A\implies x\in A$. As $A$ is open, there is an open nbhd $N$ of $x$ such that $x\in N \subseteq A$. This implies that $N\cap (\mathbb{R^n}-A)=\varnothing$.

Conversely, suppose that $\mathbb{R^n}-A$ is closed. Every point $x\notin \mathbb{R^n}-A$ has a nbhd $N$ with $N\cap (\mathbb{R^n}-A)=\varnothing$, thus $N$ lies entirely inside $A$ (i.e., $x\in N \subseteq A$). So $A$ is open.

The second statement follows from the first by substituting $\mathbb{R^n}-A$ for $A$

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Your reasoning seems solid. Note, however, that in many approaches closed sets are defined as somplementary to open sets, hence your proposition is just a definition. – TZakrevskiy Jul 20 '13 at 12:47
What's the defintion of closed subset you use? – user63181 Jul 20 '13 at 12:50
@SamiBenRomdhane $A$ is closed if every point $x\notin A$ is exterior. A point is an exterior point of $A$ if there is a nbhd $N$ of $x$ such that $N\cap A=\varnothing$ – saadtaame Jul 20 '13 at 12:54
@HagenvonEitzen But I mean that this defintion of closed subset given by the OP is almost saying it's a complement of an open subset. – user63181 Jul 20 '13 at 13:04
I fail altogether to understand the downvote: the OP has clearly invested effort in the problem. – Brian M. Scott Jul 20 '13 at 23:32

## 2 Answers

The first paragraph in your proof is absolutely right. The Second claim is most easily done by using the first one and the fact $$\Bbb R^n - \left(\Bbb R^n - A\right)=A.$$

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1) $(\implies)$ You are almost there, assume $A$ is an open set. Let $x\in A$, then we can find an open set $O$ such that $x\in O \subset A$ which implies $O \cap A^c = \phi$ since $A \cap A^c = \phi.$

Note: $A^c=R^n\setminus A$ is the complement.

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It seems to me that you’re just restating the OP’s argument. – Brian M. Scott Jul 20 '13 at 23:38
@BrianM.Scott: Not exactly. Thanks for the edit. – Mhenni Benghorbal Jul 21 '13 at 0:21
I see no significant difference, and no basis for the assertion that the OP is only almost there. – Brian M. Scott Jul 21 '13 at 0:22
@BrianM.Scott: Note that, the OP is saying $N \subseteq A$ and it should be $\subset$. – Mhenni Benghorbal Jul 21 '13 at 0:25
Eh? There’s certainly nothing wrong with $N\subseteq A$. Indeed, by any possible definition of neighborhood it is certainly possible that $A$ is a nbhd of $x$. – Brian M. Scott Jul 21 '13 at 0:33