# closed and open set - set $S$ is open if and only if its complement is closed?

Let set $S$ be a set of real numbers. A point $p∈S$ is set to be interior point of $S$ provided that there exist a $δ>0$ such that $(p-δ,p+δ)⊆S$. The set $S$ is said to be an open set if every element of $S$ is an interior point.

How can I prove that

Set $S$ is open if and only if its complement is closed.

• How do you define "closed"? Commented Apr 23, 2013 at 17:18
• As Asaf's comment suggests, closed is sometimes defined as the complement being open. We need to know what definition you have used in class for what a closed set is. Commented Apr 23, 2013 at 17:23
• Is your definition that every convergent sequence in a closed set converges to a limit point in that set? Commented Apr 23, 2013 at 17:24

Alternatively, a set $X$ is closed if every limit point of $X$ is a point of $X$. I use $S$ to represent a given set, and $S^c$ to denote the complement of $S$.

So, first suppose that $S^c$ is closed. Choose $x \in S$. Then $x\notin S^c$, and $x$ is not a limit point of $S^c$, because $S^c$ already contains all its limit points, and $x \notin S^c$. So there exists a neighborhood $N$ of $x$ such that $S^c \cap N$ is empty. That neighborhood $N$ such that $\exists \delta$ so that $N = (x - \delta, x + \delta) \subseteq S$. Thus $x$ is an interior point of $S$, and since $x$ was chosen arbitrarily, every $x \in S$ is an interior point of $S$. By definition, if every point of $S$ is an interior point of $S$, it is open. Thus $S$ is open if $S^c$ is closed.

Next, suppose $S$ is open. Since $S$ is open, every point in $S$ is an interior point of $S$. Let $x$ be any limit point of $S^c$. Then every neighborhood of $x$ contains a point in $S^c$, and so $x$ is not an interior point of $S$. Hence it must follow that that $x \in S^c$. Thus every limit point of $S^c$ is a point in $S^c$. And by the definition of a closed set, it follows that $S^c$ is closed, if $S$ is open.

• Diane - are these definitions (open set, closed set, limit point, neighborhood) familiar to you? Commented Apr 23, 2013 at 18:10
• yes, they are very familiar to me, thank you very much for your great answer. Commented Apr 23, 2013 at 20:31
• You're very welcome, Diane! Commented Apr 23, 2013 at 20:35
• @amWhy: you are a very good writer. +1 Commented Apr 24, 2013 at 0:40
• @amWhy: You won't believe how eager I am to be a good writer in English, Amy so plus plus. :-) Commented Aug 22, 2013 at 6:32

Assuming the definiton of a set $A$ being closed is that every convergent sequence in $A$ converges to a limit point in $A$, here's a rough sketch of the proof:

First off, it's easier to prove that a set $S$ is not open if and only if its compliment is not closed.

Assume $S$ is not open. Then there is some point $s \in S$ that is not an interior point of $S$. This means that for every $\delta > 0$, there is a point in $(s - \delta, s + \delta)$ that is not in $S$, hence is in the compliment of $S$. So some sequence in the compliment of $S$ converges to $s$ which is not in the compliment of $S$, hence the compliment of $S$ is not closed.

Now assume the compliment of $S$ is not closed. This means there is some sequence ${s_n}$ in $S$ that converges to $s$ not in the compliment of $S$. So $s$ is in $S$, and since ${s_n}$ converges to $s$, for every $\delta > 0$, there is some $s_n \in (s - \delta, s + \delta)$. So $s$ is not an interior point of $S$, hence $S$ is not open.

The first direction: Take a converging series in the complement. If it does not converge in the complement, than this point is no inner point in the original set.

The other direction: If the original set has a point which is no inner point, then you can construct a series in the complement, which converges to the inner point.