The set is $\{ \frac{1}{p} + \frac{1}{q} : p, q \in \mathbb{N} \setminus \{ 0\}\} \subseteq \mathbb{R}$
So, I think the set defined is a subset of $\mathbb{Q}$. The interior is empty, the closure is the set of all nonnegative reals, the boundary is the closure minus the interior so again is the set of all nonnegative reals. The limit/accumulation points are nonnegative reals. And there are no isolated points I believe.
Does anyone have comments on this? This is partly a question/discussion type of post meant to further my understanding.
Be careful with this sort of thing! I agree that the impulse is to say that the closure is the set of nonnegative reals, but pay attention to the details of the set. Intuitively, a point is in the closure if you can get "close" to it while staying inside your set. How would you get close to, say, $3$? $\frac{1}{p}$ is at most $1$, and $\frac{1}{q}$ is also at most $1$. So $\frac{1}{p} + \frac{1}{q}$ is at most $2$, which is nowhere near $3$.
That immediately tells us to be more careful, so let's think about what points we can get close to. We can get close to $\frac{1}{p}$ for any $p \in \mathbb{N} \setminus \{0\}$, by just taking $q$ larger and larger. And we can get close to $0$, by taking $p$ and $q$ both large. But we can't get close to anything else. So: the closure of this set is $\{\frac{1}{p} + \frac{1}{q} \mid p,q \in \mathbb{N} \setminus \{0\}\} \cup \{\frac{1}{p} \mid p \in \mathbb{N}\} \cup \{0\}$. This is also the boundary, because you're correct -- the interior is empty. The limit points are the extra things we just added in; $\frac{1}{p}$ for positive integers $p$, and also $0$.
There are, however, isolated points. $2$ is one of them: we can hit $2$ by taking $p = q = 1$, but increasing either $p$ or $q$ takes us at least a distance of $1/2$ away from $2$. In fact, every point in our set is like this!