Let $E^{\circ}$ denote the set of all interior points of a set $E$
Prove that $E^{\circ}$ is always open.
Important definitions:
An interior point of a set E, is a point $p$ in which there exists a nbhd $N_{\delta}(p)$ such that $N_{\delta}(p) \subset E$.
A set is said to be open if every point of $E$ is an interior point
My Solution: I would've just said that the set $E^{\circ}$ is open by this definition.
But another solution I found was:
If $x\in E^{\circ} \rightarrow x\in U \subset E^{\circ}$ where $U$ is open.
Therefore any point $x\in U$ has a nbhd $N_{\delta}(p) \subset E$.
This implies $E^{\circ} = \bigcup _{U \subset E}U$.
And from this it is concluded that $E^{\circ}$ is open.
My question is why is this extended argument necessary? This happens to me often where I encounter a question which appears to be asking something simple, but the actual proof is more complicated than initially anticipated.