In complex analysis, how do I show that a set S is open iff each point in S is an interior point.
Thank you much.
The interior of a set S is the largest open set contained in S. If S is open clearly the interior of S is S so every point of S is in the interior of S since they are the same.
Conversely if every point of S is an interior point then S \subset $\displaystyle S^{\circ}$. By definition of interior you always have the reverse containment. So since $\displaystyle S=S^{\circ}$ S is open.
The approach to this question depends upon the definitions that your textbook uses.
A text by Churchill is a fairly typical undergraduate complex analysis text.
That text uses the statement of this problem as the definition of ‘open set’.
Begin with the idea of $\displaystyle \varepsilon$-neighborhood: $\displaystyle \varepsilon > 0\; \Rightarrow \;N_\varepsilon (z_0 ) = \left\{ {z:\left| {z_0 - z} \right| < \varepsilon } \right\}$.
In almost all developments, open sets as well as interior points are defined by $\displaystyle \varepsilon$-neighborhoods.
So you need to follow the way that your textbook gives the definitions.