Closure in Metric Space

Question

I need help understanding Theorem 2.27(c) in Rudin.

If $X$ is a metric space and $E \subset X$, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $\bar {E} = E \cup E'$.

Theorem 2.27:

If $X$ is a metric space and $E \subset X$, then

(c) $\bar {E} \subset F$ for every closed set $F \subset X$ such that $ E \subset F$.

Proof given:

If $F$ is closed and $F \supset E$, then $F \supset F'$, hence $F \supset E'$. Thus $F \supset \bar {E}$.

How does one conclude that $F \supset E'$?

Answer 1

Because every limit point of $E$ is a limit point of $F$.

This may sound like "it is because it is." But we can give a quick proof from the definition. Recall that $w$ is a limit point of $A$ if every deleted $\epsilon$-ball with centre $w$ contains a point of $A$.

Thus if $w$ is a limit point of $E$, every deleted $\epsilon$-ball with centre $w$ contains a point of $E$, and therefore of $F$, since $E\subset F$. So $w$ is a limit point of $F$.

Answer 2

You have $E\subset F$ so $E'\subset F'$. But $F$ is closed, so $F'\subset F$.