I am in a complex analysis class and have been asked to prove this. I know I have to prove both ways so.
If a set is open then each point in $S$ is an interior point.
Proof: Let $S$ be an open set, then by definition the set contains none of its boundary points, then by definition the set $S$ contains no points that intersect both $S$ and points not in $S$.
Where do I go from here? Do I define a point $z_0 \in S$ and then say it must be an interior point because it cannot intersect both the interior and exterior of $S$ because then it would be a boundary point which contradicts the definition of open?
If each point in $S$ is an interior point then a set $S$ is open.
Let $z_0$ $\in S$
My definition for open is: A set is open if it contains none of its boundary points.
My definition for boundary points is: a point all of whose neighborhoods contain at least one point in S and at least one point not in S.
My definition for interior points is: a point is an interior point of the set S whenever there is some neighborhood of z that contains only points of S.
Best Answer
Your definition of open is quite strange, but anyways.
Let $S$ be an open set, then it doesn't contain any boundary points. Now take $s \in S$ we want to see that $s$ is an interior point (to show that every $s\in S$ is an interior point). So by the definition you have, we want to see that there exists a neighbourhood of $s$ such that it only contains points of $S$. Let's suppose that this neighbourhood doesn't exist, then for any neighbourhood of $s$ we have that there are points which are not in $S$ which belong to the neighbourhood. But this by definition means that $s$ is a boundary point, which contradicts the asumption that $S$ is open.
Therefore there must a exist a neighbourhood of $s$ that only contains points of $S$.
Can you do the other implication?