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.
Best Answer
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.