Closure of a set-Why is every set contained in a smallest closed set? I was reading the book Vector Calculus, Linear Algebra and Differential Forms- A Unified Approach by John and Barbara Hubbard, but I am unable to correctly understand the concept of a closure of a set.
The book defines closure as-

Definition 1.5.8 (Closure). If $A$ is a subset of $\mathbb{R}^n$, the closure of $A$ is the set of $x \in \mathbb{R}^n$ such that for all $r>0$ , $B_r(x) \cap A \neq \emptyset$

Note the "for all". To me it means that all the points in the closure must lie in A or its boundary, if not there exists an open ball of radius $\epsilon>0$ s.t its intersection with $A$ is the empty set.
In a previous paragraph it also says,

Every set is contained in a smallest closed set, called its closure.

Here is my confusion - 
Consider the set $\{\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 | x^2+y^2<9\}$ to be $A$
According to the definiton, isn't $\{\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 | x^2+y^2 \leq 2\}$ a closure of $A$. Similarly any arbritary disk of radius less than 9 fits in the definiton of closure. Clearly they do not contain all of $A$ and I can build smaller and smaller disks.
Why then did the author say the closure of $A$ is the smallest closed set to contain $A$ ? 
I seem to be misunderstanding something crucial here and I'd like to ask if anyone be kind enough to tell me where I'm going wrong.
 A: Set $\{(x,y)\in \Bbb{R^2}|x^2+y^2\leq2\}$ doesn't fit the definition of closure for $A=\{(x,y)\in \Bbb{R^2}|x^2+y^2<9\}$ because it doesn't contain all $a\in \Bbb{R^2}$ that satisfy: $(\forall r>0)\ B_r(a)\cap A \neq \emptyset$.
Closure of $A$ is set $clA=\{a\in\Bbb{R^n}|(\forall r>0) B_r(a)\cap A \neq \emptyset\}$, not $clA\subset \{a\in\Bbb{R^n}|(\forall r>0) B_r(a)\cap A \neq \emptyset\}$
This part is true: "Note the "for all". To me it means that all the points in the closure must lie in A or its boundary, if not there exists an open ball of radius $\epsilon>0$ s.t its intersection with $A$ is the empty set."
A: Let me use the shorter notation $B_r(p)$ for an open ball (open disk) of radius $r$ centered at $p$.
You are asserting that $B_2(0)$ fills the bill to be the closure of $B_9(0)$.  That's not true.  Consider the point $x = (0,3)$, which is in $B_9(0)$.  For every $r > 0$, $B_r(x)$ contains at least one point of $B_9(0)$ - for instance, $B_r(x)$ contains $x$ itself.  The closure of $B_9(0)$ was defined to be the set of all points having that property, so $x$ has to be an element of the closure.  But $x$ is not in $B_2(0)$.  So $B_2(0)$ cannot be the closure of $B_9(0)$.
This logic shows that the closure of any set has to be a superset of that set.
