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

Question

Is my proof of the following correct?

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

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
@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 $N\cap A=\varnothing$ — mrk, Jul 20 '13 at 12:54
But do you not see that $x\not \in A\iff x\in A^c$ and $N\cap A=\emptyset\iff N\subset A^c$ so your defintion of $A$ is closed is equivalent to $A^c$ is open ? — , Jul 20 '13 at 12:59
@SamiBenRomdhane Well, all valid definitionms of closed are equivalent to complement of open at the end of the day ... — Hagen von Eitzen, Jul 20 '13 at 13:00
@SamiBenRomdhane That's my proof, what do you mean 'do you not see that'? — mrk, Jul 20 '13 at 13:02
@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. — , Jul 20 '13 at 13:04
@SamiBenRomdhane It's an exercise in a book. Is my proof correct? — mrk, Jul 20 '13 at 13:05
@saadtaame Yes, it's fine. Though your second part should should formally start with $x\in A$ and not iwth the (equivalent of course) $x\notin \mathbb R^n-A$. — Hagen von Eitzen, Jul 20 '13 at 13:09
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

score 0 · Answer 1 · edited Jul 20 '13 at 23:36

0

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.

edited Jul 20 '13 at 23:36

Brian M. Scott

616,228

answered Jul 20 '13 at 13:23

Mhenni Benghorbal

47,431

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 00:21
3

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 00:22
@BrianM.Scott: Note that, the OP is saying $N \subseteq A$ and it should be $\subset$. – Mhenni Benghorbal Jul 21 '13 at 00:25
2

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 00:33

score 0 · Accepted Answer · answered Jul 21 '13 at 22:08

0

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

answered Jul 21 '13 at 22:08

leo

10,433

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

2 Answers2