If E' is the set of all limit points of a set E what does (E')' represent? The answer to this question is sure to make me feel embarrassed, but here we go:
I am wondering about notation if $E'$ is the set of limit points of a set E, then what does $\left(E'\right)'$ mean? The reason I wonder is since the book uses $E^c$ for the complement of E, hence the representation of $\left(\right)'$ must mean something different?
 A: $(E')'$ is the set of limit points of limit points of $E$ - and this is in general not the same as $E'$.
To see this, first consider the set $X=\{1, {1\over 2}, {1\over 3}, {1\over 4}, ...\}$ - clearly $X'=\{0\}$.
Now for $a, b$ positive real numbers, let ${X\over a}+b=\{{x\over a}+b: x\in X\}$ - this is just the set $X$, compressed by a factor of $a$ and then shifted to the right $b$ units. Note that $({X\over a}+b)'=\{b\}$.
We're going to place a bunch of shrinking copies of $X$ next to each other.
Let:


*

*$Y_0={X\over 1}+0$ (this is just $X$)

*$Y_1={X\over 2}+1$

*$Y_2={X\over 4}+{3\over 2}$

*$Y_3={X\over 8}+{7\over 4}$
.
.
.


*

*$Y_n={X\over 2^n}+{2^{n}-1\over 2^{n-1}}$.


Now let $Z=\bigcup Y_n$.
If you draw this, you'll see that $Z'=\{0, 1, {3\over 2}, {7\over 4}, ..., {2^n-1\over 2^{n-1}}, ...\}$. But this set itself has a limit point, namely $2$; so $(Z')'=\{2\}$.

We can go further. We can similarly define $E^{(n)}$ for each $n$: set $E^{(0)}=E$, $E^{(n+1)}=(E^{(n)})'$. We can even keep going "past infinity": we can define $E^{(\alpha)}$ for $\alpha$ an arbitrary ordinal, by expanding the definition above to include $E^{(\lambda)}=\bigcap_{\alpha<\lambda}(E^{(\alpha)})$ for $\lambda$ limit. This is called the Cantor-Bendixson process. 
Let's say we work inside $\mathbb{R}$ with the usual topology specifically. Then it turns out that for each set $X$, there is some countable ordinal $\alpha$ such that $X^{(\alpha)}=X^{(\alpha+1)}$ (that is, the hierarchy stabilizes at some countable stage), but at the same time there are sets (even closed sets!) which take arbitrarily large countable ordinals to stabilize.
One neat application of this is that every closed set in $\mathbb{R}$ can be uniquely written as the disjoint union of a countable set and a perfect (= closed with no isolated points) set. If you're interested in this sort of thing, look up descriptive set theory.
A: Let's compare $(E')'$ with $E'$. 

Claim: In general, it's neither true that $(E')' \subset E'$ nor that $E' \subset (E')'$.

Here's a somehow satisfying argument for the first claim:
Let $x \in (E')'$. Let $n \in N(x)$. We know that $n \cap (E' -\{x\}) \neq \varnothing$, so $\exists y \in n \cap E'$ such that $y\neq x$. As $y \in n$, $n \in N(y)$, so as $y \in E'$, $n \cap (E - \{y\}) \neq \varnothing$, i.e. $\exists z \in n\cap E$ with $z \neq y$. Notice that this doesn't lead to the desired result (which is $x \in E'$). Though, if the space is $T_1$, then we could've consider $n-\{x\}$ as a neighborhood of $y$, and adjusting the above argument we get a $z \in (n-\{x\}) \cap (E-\{y\})$, so we are done because $ (n-\{x\}) \cap (E-\{y\}) \subset n \cap (E-\{x\})$. 
So, in general, $(E')' \not \subset E'$ unless the space is $T_1$. 
For the other way around, consider $X = \{0,1,2\}$, $\tau = \{\varnothing, \{1,2\},X\}$. For $E = \{0,1\}$, $E' = \{0\}$, and $E'' = \varnothing$. 
