Limit points of S = $\{\frac{1}{n}, n \in \mathbb{N}\}$, what about irrationals? I've been thinking about limit points and irrationals. Why is there not an irrational just a bit bigger than zero such that it is not in the set S and for every $\delta > 0$ it contains some element of S? For that matter why are there not irrationals between every $\frac{1}{n}$ that act as limit points?
 A: Notice that 
$$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$$
You can certainly take any irrational between these two numbers and choose a $\delta > 0$ small enough so that the ball with radius $\delta$ doesn't contain either of those two numbers. Thus, if you can find one open set that contains the potential limit point, but does not intersect the set, that's enough to show that it's not a limit point. Hopefully this answers both of your questions. 
A: Let us imagine such an irrational $w$, as you say close to $0$ (or at least less than $1$) and bigger than $0$.  Then there is a natural number $k$ such that $\frac{1}{k}\lt w$. 
The fact that there is such an $k$ is called the Archimedean Property of the reals. Informally, it says that there are no infinitesimally small non-zero quantities.  We omit the proof that the reals have the Archimedean property, but can sketch a proof on request. One of the usual proofs uses the least upper bound property.
Let $N$ be the largest natural number such that $\frac{1}{N}\le w$. By our assumptions, $N\gt 1$ and $w$ is strictly between $\frac{1}{N}$ and $\frac{1}{N-1}$. 
Let $\epsilon=\min(w-\frac{1}{N}, \frac{1}{N-1}-w)$. Then for all integers $k$, we have $\left|w-\frac{1}{k}\right|\ge \epsilon$. This shows that $w$ cannot be a limit point of $S$.
