Is my notion of boundary and bounded correct?
Boundary as I understand it is the set of all points for with we can center a ball B(x,r) with radius r>0 with contains points that are both inside and outside the set.
Bounded means there is a certain number C for which is bigger then every point in our set. ie ||x|| < C.
So my question is the following: why does a bounded set need not have a boundary ? Shouldn't one imply the other?
On
Take the following counterexample: on the Euclidean plane, the open ball of radius $1$ centered in $(0,0)$ is a bounded set, because each point $\mathbf{x}$ in it obeys to the relation $\lVert \mathbf{x} \rVert \le 1$. However, its boundary is the empty set: because $\lVert \mathbf{x} \rVert \le 1$, you can assume that $$ \lVert \mathbf{x} \rVert = 1 - \epsilon, \quad \epsilon \in (0,1]. $$ Thus, there are no points which cannot be included in a open ball wholly contained in the set: for $\mathbf{x}$, the open ball centered in it of radius $\frac{\epsilon}{2}$ is always fully contained in the set.
The boundary of a set is made of all the points with the following condition: every open ball centered in $\mathbf{x}$ both intersect the set and its complement.
The boundary of a subset $S$ of a topological space is the set of points in the closure which are not in the interior, i.e. $\partial S=\overline{S}\setminus S^\circ$. It is a simple exercise using the definition of boundary point given in your question to prove this.
If $S\subset\mathbb{R}^n$ is nonempty and bounded, then we have both $\overline{S}\neq\emptyset$ and $\overline{S}\neq\mathbb{R}^n$, so that since $\mathbb{R}^n$ is connected, $\overline{S}$ cannot be both open and closed. If $\partial S=\emptyset$, then this would imply that $\overline{S}=S^\circ$. But this is impossible, since $\overline{S}$ is closed but $S^\circ$ is open.
In fact, any normed linear space is connected (use that path-connected implies connected), so this same argument shows that in any normed linear space, a nonempty bounded set has nonempty boundary.