# The definition of the closure of a set

Let $A\subseteq\mathbb{C}$ be given. The closure of A is $\overline{A}=\left\{z:\forall\epsilon>0, D_{\epsilon}\left(z\right)\cap A\neq\phi\right\}$

I was wondering whether someone could expand on/explain the above definition as I'm not quite sure what it means. I understand that $D_{\epsilon}\left(z\right)$ is the open disk with centre $z$ and radius $\epsilon$ and that A is a set.

$\bar A$ is the set of all complex numbers $z$ with the property that given any real number $\varepsilon>0$, however small, any disc with centre $z$ and radius $\varepsilon$ must contain at least one element of $A$.
If for every $\varepsilon>0$ $D_\varepsilon(z)$ contains at least one of element of $A$ other than $z$ itself, $z$ is called a limit point of $A$. Note that a limit point of $A$ need not belong to $A$ at all. For example, if $A=\{x+0i:x\in\mathbb R, 0<x<1\}$, then $0$ is a limit point of $A$ even though $0\notin A$. On the other hand a point can belong to $A$ and yet not be a limit point of $A$. For example, if $A=\{z\in\mathbb C:|z|=1\}\cup\{0\}$ then $0$ is not a limit point of $A$ even though $0\in A$.
If $A'$ denotes the set of all limit points of $A$, then the closure of $A$ is $\bar A=A\cup A'$. This is the smallest closed subset of $\mathbb C$ containing $A$.
The closure of $A$ is the set obtained by adding the limit points of $A$ to $A$. In other words, it is the set formed by the elements of $A$, along with the elements of your topological space such that they do not have any neighborhoods disjoint from $A$. In a metric space such as $\mathbb{C}$, this definition can be refined by saying that $x$ is a limit point of $A$ if every open ball around $x$ also contains some point of $A$. It is obvious that every element of $A$ is a limit point of $A$, so we may define the closure of $A$ as the set of all of its limit points.
For example, consider the set $S = \{ z \in \mathbb{C}: |z| < 1 \}$. $1$ is a limit point of this set, because if we have any open ball centered at $1$, this ball will include an element of the form $1 - \delta$ for $\delta > 0$, which will then be in $S$. Likewise, any element such that $|z| = 1$ is a limit point, and if $|z| > 1$ the open ball $D_{\epsilon}(z)$ where $\epsilon = (|z| - 1)/2$ is disjoint from $A$, so $z$ is not a limit point. This means that the closure of $S$ is $\{ z\in \mathbb{C} : |z| \leq 1 \}$.