Closure of a subset of a metric space is closed From definition, if $X$ is a metric space, if $E \subset X$, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $\overline{E}=E \cup E'$.
I need to prove that $\overline{E}$ is closed. By a closed set, I mean that all limit points of a set are in the same set. A limit point of a set is a point whereby every neighbourhood of the point contains a $q$ such that $q$ is an element of the set.
The proof provided in Theorem 2.27(a) of Rudin's Principles of Mathematical Analysis is as follows:
If $p \in X$ and $p \notin \overline{E}$ then $p$ is neither a point of $E$ nor a limit point of $E$. Hence $p$ has a neighbourhood which does not intersect $E$. The complement of $\overline{E}$ is therefore open. Hence $\overline{E}$ is closed.
My question is with how "Hence $p$ has a neighbourhood which does not intersect $E$." leads to "The complement of $\overline{E}$ is therefore open.".
To show that the complement of $\overline{E}$ is open, we need to show that $p$ has a neighbouthood which does not intersect $\overline{E}$, not just $E$. Unless it is true that if the neighbourhood (an open set, since neighbourhoods are all open sets) does not intersect $E$, then it also does not intersect $E'$. Is that true?
 A: Your definition of limit point is a little off: it should say that $p$ is a limit point of a set $E$ if every nbhd of $p$ contains a point $q\in E$ such that $q\ne p$. Without that last condition $2$ would be a limit point of $[0,1]\cup\{2\}$ in $\Bbb R$, which is not what we want the term to mean.
You’ve started with a point $p\in X\setminus\overline E$, meaning that $p\notin E$ and $p\notin E'$. Since $p\notin E'$, $p$ has a nbhd $U$ that contains no point of $E$ different from $p$. Since $p\notin E$, this means that $U$ contains no point of $E$ at all: $U\cap E=\varnothing$. Suppose that $x\in U\cap E'$; then $U$ is a nbhd of $x$, and $x\in E'$, so $U\cap E\ne\varnothing$, which we just saw is not the case. Thus, $U\cap E'=\varnothing$, and therefore $U\cap\overline E=\varnothing$, as desired.
A: In a metric space, a set $E$ is open if and only if $\forall x \in E$ there is an open ball around $x$ contained in E (by definition), i.e. $B(x,\delta)\in E$. Such an open ball is a neighborhood of x.
In your case, $\forall p \in \bar E\complement$ we have found an open neighbourhood of $p$ in $\bar E\complement$, which contains an open ball of $p$ contained in $\bar E$
Note that, if $p \in E\complement$ has a neigbourhood which does not intersect $E$, then it cannot intersect $\bar E$. If it did, then we would have a point $x \in \bar E \setminus E$ with $x \in E\complement$, and an open neigbourhood around $x$ in $E\complement$. Thus $x$ could not be a limit point of $E$, and hence is not in $\bar E$, contradiction.
A: Assume that you have in that neighborhood V a point from $E'$, then this point is a limit of a series from $E$ and therefore in each neighborhood, including V, which is a contradiction.
A: Remember that a limit point of $E$ is defined to be a point $x \in X$ such that in every neighborhood of $x$, call it $N(x)$ there exists a point $y \in N(x) \setminus \{x\}$ so that $y \in E$.
Inherently since we have some point $p$ with some neighborhood $N(p)$ which doesn't intersect $E$. Let's assume it intersects $E'$. This means there's some point $q \in N(p)$ so that every neighborhood of $q$ intersects $E$. Now topologically speaking $N(p)$ is also a neighborhood of $q$ which would mean that $N(p)$ intersects $E$, a contradiction. However Rudin defines a neighborhood of a point as an open ball centered at that said point, so let's work through this further. Let's take some neighborhood of $q$, $N(q)$ which is a strict subset of $N(p)$ (why can we do this?). Now since $q$ is a limit point we know that $N(q)$ intersects $E$, but that would mean that $N(p)$ also intersects $E$ (since $N(q) \subsetneq N(p)$), a contradiction to the fact that $N(p)$ doesn't intersect $E$. Thus now we know that $N(p)$ doesn't intersect $E'$ either, thus $N(p)$ doesn't intersect $\overline{E}$ so that we conclude that $\overline{E}$ is closed.
