Show that in a discrete metric space, every subset is both open and closed. I need to prove that in a discrete metric space, every subset is both open and closed. Now, I find it difficult to imagine what this space looks like. I think it 
consists of all sequences containing ones and zeros. 
Now in order to prove that every subset is open, my books says that for 
$A \subset X $,
$A$ is open if $\,\forall x \in A,\,\exists\, \epsilon > 0$ such that $B_\epsilon(x) \subset A$.
I was thinking that since $A$ will also contain only zeros and ones, it must be open. Could someone help me ironing out the details? 

In a discrete space, take $\epsilon=\frac{1}{2}$. Then the ball contains only the point itself so that it is a subset of $A$. Every subset is closed because the complement of an open set is closed.
