Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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.
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.
