Let $E$ be a set in a metric space $X$ and $E'$ denote the set of all limit points and $\bar{E}$ be the closure of $E$. Then $\bar{E} = E \cup E'$
Prove that $E'$ is closed, and $\bar{E}$ and $E$ have the same limit points
So as sets, they look like?
I know that the $p'$ just solely inside $E'$ is wrong, but is the actual picture even right?
I read a lemma that since $E'$ is the set of all limit points, any point (say) $p'$ must contain another point in $E$ for that point isn't the point itself