Where am I wrong with my reasoning in Closed or Open Sets? Why the word "around" - in the following sentence - does not imply that there must be something outside of A that will be in the ball?
I saw this sentence here (Bounded vs. unbounded, closed vs. open sets), a very good explanation of Bounded vs. Unbounded and Open vs. Closed Sets.
However, I'm in the middle of learning Real Analysis and these types of sentences confuse me a lot because even though I understand the statements and the logical structures of the mathematical definitions (such as the one about Open Sets) it is very hard for me sometimes to accept something like: 
-> "I'm putting an open ball around any point of A and all points in the tiny ball will be in A"
So, here it goes my question and I would like a mathematical answer as well as a didactic way to transform my thinking so that I do not trick myself into thinking more than I should:
-> Okay, so if you drew a tiny ball AROUND any point of A you are probably able to draw a tiny ball AROUND all points of A. It does not matter to me if A is infinite because you just told me you are able to draw a tiny ball AROUND ANY POINT of A. Thus, how come you will not end up intersecting whatever it is that lies outside of A (since A is real, what lies outside of A has to be real numbers as well!)? That is, why can't I think that there is something that lies outside of A and that these elements will be intersected by this tiny ball? Even if you tell me that you can change the radius of this ball how many times you want (that's why we always say "for all positive real epsilon") I will still say that you can and will intersect something outside of A in case you are realy drawing balls AROUND ANY point of A.
I hope I was clear about my concern on the way I usually think when I read such a definition. Why I'm I in trouble when I think this way and why I am wrong?
Thank you!
 A: This is the actual quote from fleablood's answer:

A set $A$ is open: if for every point $a$ of $A$ we can draw a tiny ball around $a$ and all the points in the tiny ball will be in $A$.

Choose a single point $a$ from $A$. Consider if you can draw a tiny ball $B$ such that $a \in B \subseteq A$ ($B$ contains the point $a$, and $B$ lies entirely inside $A$).
If this is possible to do for any point $a$ from $A$, then $A$ is open. The important thing to note is that the tiny ball $B$ will be different for different points $a$. For instance, if $a$ is close to a boundary of $A$, then naturally $B$ must have a very small radius in order to not have points outside of $A$.
A: Just because you can draw a tiny ball around any point in A, you cannot draw a tiny ball around all of A. Suppose A is the open unit disk in $\mathbb{R}^2$ ($\{x: |x-(0,0)| <1\}$ (where $|\cdot|$ represents the usual norm in $\mathbb{R}^2$ ) . Choose a point (say the origin). You can draw a tiny (radius epsilon) circle around the origin. A single tiny circle cannot encompass all of A, as the radius of A is 1 and the radius of a tiny circle is smaller than 1 (when we say tiny, we mean that we can make it as small as we want). 
Your sentence of choice says that if you pick any point in $A$ (say $A$ the unit disk and think of the point (.99,.99)). Then if you make a small enough circle (in this case radius .001 is  small enough) centered at your point, it remains in the set. Every point in this circle I just chose is contained in $A$ (the unit disk).  
Constructing examples should help. 
