I want to determine the box-counting dimension (Minkowski Dimension) of the set $S=\{\frac{1}{n^2} \ | \ n \in \mathbb{Z}^+\}$. My first attempt was to first define my $\delta_n$. What I did was consider the distance between two points by
$$\frac{1}{n^2} - \frac{1}{(n+1)^2} =\frac{2n+1}{n^2(n+1)^2}$$
So I defined $F = \left[\frac{1}{(n+1)^2}, \frac{1}{n^2}\right]$ to be my covering set with diameter $\delta_n = \frac{2n+1}{n^2(n+1)^2}$. Now, I decided that because this is a completely disconnected set, I can use a single interval for each point prior to the $n$ that I am at. So if $n=4$, then I only need 3 of $F$ to cover $1,1/4,1/9$. Then I decided I can cover the all other points by finding how many of $F$ would cover all the points from $0$ to $\frac{1}{n^2}$. So that should just be $\frac{n^2}{2n+1}$ of them. Thus, $N_\delta(F) = n+\frac{n^2}{2n+1}$. Then I took the limit
$$\lim_{n\to\infty} \frac{log\left(n+\frac{n^2}{2n+1}\right)}{-log\left(\frac{2n+1}{n^2(n+1)^2}\right)}$$
Then using Mathematica, I got that this goes to 1. Is this right? Why would this disconnected set have dimension 1? Does it have to do with the closure? I have also see that the set $S=\{\frac{1}{n} \ | \ n \in \mathbb{Z}^+\}$ has box-counting dimension 1/2. Any advice?
You need to use a uniform grid. It’s easiest if you take the interval length to be $\frac1{n^2}$; then $\left[0,\frac1{n^2}\right]$ covers all but the $n-1$ points $\frac1{k^2}$ with $1\le k<n$, each of which requires its own interval (since no interval of length $\frac1{n^2}$ contains two of these points), so we end up looking at
$$\frac{\ln n}{\ln n^2}=\frac{\ln n}{2\ln n}=\frac12\,.$$
For intervals of arbitary length $\epsilon>0$ it’s more difficult to count the boxes needed to cover the set, but it’s not hard to see that if $\frac1{(n+1)^2}\le\epsilon\le\frac1{n^2}$, the number of $\epsilon$-boxes needed must be either $n$ or $n+1$, so the limit must be $\frac12$.