Classic problem from Baby Rudin that I want to see if I proved correctly:
Is every point of every open set $E \subset \mathbb{R^2}$ a limit point of $E$ ?
Solution:
Suppose $E$ is open and there exists a point $p \in E$ that is not a limit point.
Since $p$ is not a limit point,
$\Rightarrow$ there exists nbhd $N_r(p)$ of set $E$ such thtat for all $q \neq p$, $q \notin E$
But given that $E$ is open, this means $p$ is an interior point of $E$.
$\Rightarrow$ there exists a nbhds $N_\epsilon(q) \subset N_\delta(p) \subset E$ such that $q \in E$ and $q\neq p$
if we take min{$r, \delta$} we have a ball around $p$ where $N_\epsilon(q) \subset B_{min(r,\delta)}(p)$
$\Rightarrow$ there exists points $q \in E$ ,$q\neq p$, and $ q\in B_{min(r,\delta)}(p)$
This is a contradiction because $p$ was assumed to be a limit point
Q.E.D
Best Answer
It doesn't seem like you have shown a contradiction. It seems like you have shown that there exists a neighborhood $N_r(p)$ such that all points in that neighborhood (other than $p$ itself) are outside $E$, and there exists a neighborhood $N_\delta(p)$ such that some point in the neighborhood is in $E$. But the two neighborhoods need not be the same, so what is the contradiction?
You're very close though.