I have a homework that says the following:
If $N_\delta(p)$ is a neighborhood of $p$ that does not intersect $p,$ show that it cannot intersect $E'$.
where E is a subset of R and where the set made by all the accumulation points of E is called the derivative set and it is indicated with E′
Any help in this proof would be appreciated.Thank you this is my attempt
I think there is a typo in your problem set. I believe this supposed to read "Let $N_\delta(p)$ be a neighborhood of $p$ that doesn't intersect $E$. Show that it cannot intersect $E'$."
We'll prove the contrapositive: Suppose $N_\delta(p) \cap E' \neq \emptyset$. So there is some $x \in N_\delta(p)$, such that $x$ is a limit point of $E$. Let $\epsilon < \delta - d(x,p)$. Since $x$ is a limit point of $E$, there is some $e \in E$ such that $e \in N_\epsilon(x)$. Can you take it from here?
WHY I THINK THIS IS THE CORRECT INTERPRETATION:
OP looks to be taking a first course in real analysis, which grounds the level of problem difficulty to be expected. Any neighborhood of $p$ must intersect $p$ (as I interpret this), so the problem must lie there. Note that this must mean $p \not \in E$, which would explain why the problem doesn't state $p \in E$.