# $A$ is closed if and only if $\mathbb{C}\setminus A$ is an open

Let $A\subset\mathbb{C}$. Prove that $A$ is closed if and only if $\mathbb{C}\setminus A$ is an open subset of $\mathbb{C}$.

Let $A$ be closed. Suppose $\mathbb{C}\setminus A$ is not open. That is, $\mathbb{C}\setminus A$ does not contain all its interior points. Then there exists a $w\in\mathbb{C}\setminus A$ such that $D(w,r)\not\subseteq \mathbb{C}\setminus A$ for any $r>0$. Or, equivalently, $D(w,r)\cap A\not=\emptyset$. Thus, $w$ is not an exterior point of $A$. Since $w\notin A$, $w$ is not an interior point of $A$. It follows that $w$ is not a boundary point of $A$, yielding a contradiction. Thus $\mathbb{C}\setminus A$ is open.

Conversely, let $\mathbb{C}\setminus A$ be open, then suppose $A$ is not closed. Then there exists a boundary point $w$ of $A$ which is not an element of $A$, hence $w\in\mathbb{C}\setminus A$. We know $w$ is not an exterior point of $A$, so for any $r>0$, we have that $D(w,r)\cap A\not=\emptyset$. That is, for any $r>0$ we have $D(w,r)\not\subset\mathbb{C}\setminus A$. Thus, $w\in\mathbb{C}\setminus A$ is not an interior point of $\mathbb{C}\setminus A$, hence $\mathbb{C}\setminus A$ is not an open subset, which is a contradiction. Hence $\mathbb{C}\setminus A$ is open.

In the first paragraph, we have shown that $w$ is neither an interior nor an exterior point. Then my lecturer states "it follows that $w$ is not a boundary point of $A$". This seems to be the exact opposite to the definition he has given us for a boundary point, namely: "A boundary point of the set $A$ is a point which is neither an interior nor exterior point".

Am I confused or is this a mistake? If so, can someone correct it using similar notation.

It should have said: yielding that $w$ is a boundary point of $A$, which is a contradiction (as $w \notin A$ and $A$ is closed so all boundary points of $A$ should already be in $A$).