Simple question about discrete metric and openness. You may think this is silly question, but I'm really confused.
In discrete metric, every singleton is an open set.
And, the proof goes like this
$\forall x \in X$, by choosing $\epsilon < 1$, $N_\epsilon(x) \subset ${$x$}
However, if discrete metric is defined in $X$,
How can we use the radius less than 1 when we only have 0 and 1?  

You don't need to take $\varepsilon<1$, you may take $\varepsilon=1$ also (can you see the reason?). But you need to know that the range of the metric (here $d$) is $\mathbb{R}^+$.
