Showing the limit points of E are equal to the limit points of E closure.

Question

I am looking at a proof to the above statement that goes as follows:

Let p $\in E'$, then by definition, any neighborhood of p contains a element q such that q $\neq$ p and q $\in E$.

And because we know $E \subset \overline{\rm E} $ (being the closure of E), this implies that q $\in$ $\overline{\rm E}$ which implies p $\in \overline{\rm E}'$ (being the closure of the set of limit points of E).

The final statement of this part of the proof is that this implies $E' \subset \overline{\rm E}'$.

My question is how does p $\in E'$ and p $\in \overline{\rm E}'$ imply that $E'$ is a subset of $\overline{\rm E}'$? It seems just as clear that you could say the other way around as well?

Answer 1

This is true by the definition of a subset.

Let $A$ and $B$ be sets. We say that $A$ is a subset of $B$, if for every element $a\in A$, the same element $a\in B$. This is denoted as $A\subseteq B$.

Your example considers an arbitrary element $p\in E'$, and then shows that $p\in \overline{E}'$. Thus, every element in $E'$ is in $\overline{E}'$, and $E'\subseteq \overline{E}'$.