$p$ is a limit point implies $p\in\overline{E-\{p\}}$. Let $E$ be a set, with $p$ as a limit point. I wish to show that if $p$ is a limit point, $p\in\overline{E-\\{p\\}}$. Denote $F=E-\\{p\\}$. The converse was proved in class, but for this direction, the proof goes: Want to show $p\in\overline{F}=F\cup \\{\text{all limit points of F}\\}=F\cup F^{\prime}$ Since $p\not\in F$ so want to show $p\in F^{\prime}$. Take each neighbourhood of $p$ to be $B_r(p)$ with radius $r$. Then $$B_r(p)\cap (F-\\{p\\} )=\emptyset$$ $$B_r(p)\cap (E-\\{p\\} )=\emptyset$$ My question is why is the intersection of the ball with $F-\\{p\\}$ empty? Isn't because $p$ is a limit point (assumption) that any neighborhood intersected with $B_r(p)\cap E-\\{p\\}\ne\emptyset$ ?

Can you show that $F\cap(B_r(p)\smallsetminus\\{p\\})=B_r(p)\cap(F\smallsetminus\\{p\\}) ?$

Then it remains you use the (equivalent) definition of closure:

> A point $x$ is in the closure of $A$ if and only if every neighborhood $N$ of $x$ intersects $A$.