Prove that the box dimension of $\{0,1,\frac{1}{2},\frac{1}{3},...\} $is$ \frac{1}{2}$ I'm supposed to consider the difference $\frac{1}{n+1}-\frac{1}{n}$ and let it equal to $\epsilon$. Hence $\epsilon=\frac{1}{n(n+1)}$. But how do I show that the number of boxes of size $\epsilon$ to cover the set is $N(\epsilon)=2n$?
After that, the proof is easy.
Thanks
edit: An easy upper bound is $N(\epsilon)\leq\frac{1}{\epsilon}=n(n+1)$ since it covers the whole interval $[0,1]$, but I don't know h0w to bring that down to $2n$.
 A: Use $n$ boxes to cover $[0,\frac1{n+1}]$.  Then use the other $n$ boxes to cover each $\frac1k$ for $1 \le k \le n$.  This shows the upper bound is $2n$.
You know that you need at least $n$ boxes of this size to cover the numbers $1/k$ for $1 \le k \le n$.  Because there is no box of this size that will cover two of these points.  This gives you a lower bound of $n$.
A: I shall use the method presented in this post. Let $F = \{0,1,1/2,1/3,...\}$ and let $|A|$ denote the diameter of set $A$.
Let $0 < \delta< \frac12$, and let $k$ be the integer satisfying $1/(k − 1)k >  ⩾ 1/k(k + 1).$ If $|U| ⩽ $, then $U$ can cover at most one of the points $\{1, 1/2 ,..., 1∕k\}$ since the distance between any pair of these points is at least $1/(k − 1) − 1/k =
1/(k − 1)k > $. Thus, at least $k$ sets of diameter $$ are required to cover $F$, so
$N_(F) \ge k$ giving
$$\frac{\log N_(F)}{-\log\delta} \ge \frac{\log k}{\log k(k+1)} \stackrel{k\to\infty}{\longrightarrow} \frac{1}{2}$$
so $\underline\dim_B F \ge \frac12$.
On the other, let $0 < \delta< \frac12$, and let $k$ be the integer satisfying $1/(k − 1)k >  ⩾ 1/k(k + 1)$. Then $(k + 1)$ intervals of length $$ cover $[0, 1/k]$, leaving $k − 1$ points of $F$ which can be covered by another $k − 1$ intervals. Thus, $N_(F) ⩽ 2k$, so
$$\frac{\log N_(F)}{-\log\delta} \le \frac{\log 2k}{\log k(k-1)} \stackrel{k\to\infty}{\longrightarrow} \frac{1}{2}$$
so $\overline\dim_B F \le \frac12$.
So we have $$\dim_B F = \frac12$$ as desired.
