Identify the boundary of $\mathbb{Q}$ in the discrete metric space $\mathbb{R}$ Consider the metric space $(\mathbb{R}, \delta)$ where $\delta$ is the discrete metric.
In other words, say that for all $x,y \in \mathbb{R}$,
if $x = y$, then $\delta(x,y) = 0$
if $x \neq y$, then $\delta(x,y) = 1$
I'm trying to identify the boundary of $\mathbb{Q}$ in $(\mathbb{R}, \delta)$
I have heard that the boundary of a discrete metric space is the empty set, but have not seen any proof for this.
I have also heard that in the discrete metric, all subsets are open (the unions of singleton sets) and hence are also closed. This does seem to lead to the boundary of the discrete metric being $A - A = \varnothing$.
Am I looking at this the right way? Is there a better way to construct a proof?
 A: I'm not sure how much elaboration is needed, so I intend to be detailed.
Following @Randall s point, as you said every set in a discrete metric space every set is open, because of the reason you mentioned; and therefore every set is complement of an open set and therefore is closed. So $ \overline{A} = A $ and $\overline{X \setminus A} = X \setminus A$. The definition $ \partial A = \overline{X \setminus A} \cap \overline{A}  $ is equivalent to yours (you can check this). And therefore for any $A$ in a discrete metric space we would have $ \partial A = A \cap (X \setminus A) = \emptyset $.
To use your definition more directly, here's another proof: Let $ x \in X$ be any candidate for being a member of $ \partial A $. Consider the open ball
$ B(x, \frac{1}{2}) = \{ y \in X \mid \delta(x,y)<\frac{1}{2} \}$. For $x$ to be in $ \partial A $, it is necessary that $ B(x, \frac{1}{2}) \cap A \neq \emptyset $ and $ B(x, \frac{1}{2}) \cap (X \setminus A) \neq \emptyset$. But in the discrete metric space we know that $ B(x, \frac{1}{2}) = \{x\}$, because every distinct point from $x$ is 1 unit away from it. Now $x$ itself is either a member of $A$ or $X \setminus A$. In both cases the open ball $ B(x, \frac{1}{2}) $ does not contain a member of both $A$ and $ X \setminus A$. And therefore $x$ cannot be a member of $\partial A$. So the proof is accomplished.
