How to get an open ball in $[0,1]$ that contains $[0,1]$? The definition of bounded we have is that if $X$ is a metric space, $z \in X$, and $X \subseteq X$, then there exists an open ball $B_z(R)$ with finite radius $R$ of $X$ centered at $z$ such that $X \subseteq B_z(R)$.
If $X = [0,1]$, it seems like the largest open ball possible is $(0,1)$ which does not contain $0$ and $1$ so by our definition $[0,1]$ is not bounded, but it obviously is. How can I get $[0,1]$ to be bounded by our definition?


If $X=[0,1]$, it seems like the largest open ball possible is $(0,1)$ …

And this is what is false. Any open ball of any radius in a metric space $X$ is by definition considered, well, as an open ball in it, regardless of how the ball or $X$ itself look like to us.
The definition of an open ball $B_z(R)$ in a metric space $X = (X,d)$ is 
$$B_z(R) = \{x ∈ X;~d(x,z) < R\}.$$
(Although most people swap the places of $z$ and $R$ and write “$B_R(z)$” instead.)
Note that by definition, the ball only consist of points “$x ∈ X$” with a certain property.
Thus, the $2$-ball at $0$ in $X = [0..1]$ e.g. is by definition $B_0(R) = \{x ∈ [0..1];~|x-0| < 2\}$. What points $x ∈ [0..1]$ satisfy $|x| < 2$?
