Thread: Open sets and sets of interior points

Need verification of a proof please

Question
Let$\displaystyle E^o$ be the set of all interior points of a set $\displaystyle E$.
Prove $\displaystyle E^o$ is always open

Proof
Consider $\displaystyle p\in{E^o}$. $\displaystyle \exists$ a neighborhood $\displaystyle N$ of $\displaystyle p$ such that $\displaystyle N\subset{E}$

This $\displaystyle N$ is an open set, so $\displaystyle \forall{q\in{N}}, \exists$ a neighborhood $\displaystyle M$ of $\displaystyle q$ so $\displaystyle M\subset{N\subset{E}}$
So $\displaystyle N$ consists of interior points of $\displaystyle E$, so $\displaystyle E^o$ is open.
QED

Proof correct? Any alternatives or ways to make it cleaner/more elegant?

Originally Posted by I-Think

Question
Let$\displaystyle E^o$ be the set of all interior points of a set $\displaystyle E$.
Prove $\displaystyle E^o$ is always open
Proof
Consider $\displaystyle p\in{E^o}$. $\displaystyle \exists$ a neighborhood $\displaystyle N$ of $\displaystyle p$ such that $\displaystyle N\subset{E}$
This $\displaystyle N$ is an open set, so $\displaystyle \forall{q\in{N}}, \exists$ a neighborhood $\displaystyle M$ of $\displaystyle q$ so $\displaystyle M\subset{N\subset{E}}$
So $\displaystyle N$ consists of interior points of $\displaystyle E$, so $\displaystyle E^o$ is open.
QED

It is correct. Basically you are noting that a neighborhood is a neighborhood of each of its points.