Can someone please verify this proof of mine relating limit points and closure of sets?

Question

Let $E$ be any subset of a metric space, let $E'$ be the set of all limit points of the set $E$, and let $E^C$ be the closure of the set $E$, i.e. the smallest closed set containing $E$. The problem ask to prove the following:

" $E^C$ and $E$ has the same limit points "

My attempt

I) If $p$ is a limit point of $E$ then $p$ is clearly a limit point of $E^C$ since $E^C = E \cup E'$. Now, if we prove that every limit point of $E^C$ is a limit point of $E$ we're done. So, suppose $p$ is a limit point of $E^C$ then every neighborhood of $p$ contains either a point of $E$ or a limit point of $E$, in the first case we're done. Let's assume the second case, let $N_r(p)$ be a neighborhood a $p$ of radius $r$ containing the limit point $q$ of $E$ then we can always find a small enough neighborhood of $q$ contained in $N_r(p)$ that contains a point of $E$ thus $p$ is a limit point of $E$ as desired.

This picture may help visualize the last step of the proof

Here, the bigger circle represents the neighborhood of $p$ while the smaller circle represents the neighborhood of the limit point $q$.

Is this proof correct? Any suggestion, tip or recommendation is welcome.

Thanks in advance.

Answer 1

Here is my modification of your proof:

Now, if we prove that every limit point of $E^C$ is a limit point of $E$ we're done. So, suppose $p$ is a limit point of $E^C$ then every neighborhood of $p$ contains either a point of $E$ or a limit point of $E$ (other than $p$ itself), in the first case we're done.

Let's assume the second case and say $q$ the limit point of $E$ such that $q\neq p$. Let $N_r(p)$ be a neighborhood of $p$ of radius $r$ containing the limit point $q$ of $E$ then we can always find a small enough neighborhood of $ q $ contained in $N_r(p)$ but not containing $p$. Since $q$ is a limit point of $E$, the small neighborhood of $q$ contains a point of $E$ which is, of course, not $p$.

Thus $p$ is a limit point of $ E$ as desired.