My proof is:
Let $x$ be any point in $\mathbb{N}$, and let $\beta$ be the closest number (such as a factor or irrational number) to $x$, such that $\beta \notin \mathbb{N}$ and $\beta > x$.
Let $B_\epsilon(x)$ be the epsilon ball around $x$ in $\mathbb{N}$. Let $\epsilon = \beta – x$.
Clearly $\forall x \in \mathbb{N}: \exists \epsilon >0, \epsilon \in \mathbb{R}: B_\epsilon \left({x}\right) \subseteq \mathbb{N}$. So we have that all singleton sets in $\mathbb{N}$ are open.
All subsets of $\mathbb{N}$, except the empty set, consist of a union of many such $x \in \mathbb{N}$ and since all unions of open sets are open, we have that all subsets of $\mathbb{N}$ are open.
This means that the complement to every subset in $\mathbb{N}$ is closed. However, since the complement to any subset of $\mathbb{N}$ is itself a subset of the space, we have that all subsets of $\mathbb{N}$ are closed.
Therefore, all subsets of $\mathbb{N}$ are clopen.
Is this correct?
Best Answer
Just show all singletons are open, which is easy, as all natural numbers are at least 1 apart: $\{n\} = B_\frac{1}{2}(n)$ where the ball is taken in the restricted metric. If all singletons are open, all subsets $A \subseteq \mathbb{N}$ are open, as $A = \cup\{\{n\}: n \in A\}$ is a union of open sets, so open.
This also makes all subsets closed, as all complements are open as well.