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Tried to prove that the following property related to Hausdorff dimension holds: $H_{dim}([0,1]) = 1$, where:

  • $H_{dim}(A):=inf\{s\in [0,+\infty] : H^{s}(A)=0 \}$ ,
  • $H^{s}(A):= \lim\limits_{\delta \to 0 }H_{\delta}^{s}(A) $
  • $H_{\delta}^{s}:=inf\{\alpha(s)\sum\limits_{i=1}^{+\infty} (\frac{diamC_{i}}{2})^{s}: A \subseteq \bigcup\limits_{i=1}^{+\infty}C_{i}, diamC_{i} \leq \delta \} $
  • $\alpha(s):= \frac{\sqrt\pi^{s}}{\Gamma(\frac{s}{2}+1)}$

Tried to derive this proof by the use of proof of contradiction by deriving contradiction in the cases $s\in [0,1)$ and $s\in (1,+\infty)$ but I failed to obtain any results.

Help in deriving this proof very appreciated!

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  • $\begingroup$ I wonder why this is tagged "contest-math"? $\endgroup$ – Mark McClure Jan 28 '17 at 21:15
  • $\begingroup$ I am preparing for Putnam. That is why. $\endgroup$ – user410985 Jan 28 '17 at 21:15
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Not sure if that is the easiest way: to calculate $H^s_{\delta}$, it suffices to consider those $\{ \widetilde C_i\}$ which are all intervals and are pairwise disjoint: If $\{ C_i\}$ is any covering of $[0,1]$. Let $\widetilde C_1$ be the smallest interval containing $C_1$. Then $\widetilde C_1$ and $C_1$ has the same diameter. Let $C^1_i = C_i\setminus \widetilde C_1$ and do the same as $n=2$: Let $\widetilde C_2$ be the smallest intervals containing $C^1_2$.

Then $\{\widetilde C_i\}$ will be covering $[0,1]$ and pairwise disjoint (possibly not, but they intersect only at the endpoints) and $\text{diam} (\widetilde C_i) \le \text{diam} (C_i)$.

Not you are essentially finding up to constant

$$H_\delta^s([0,1]) = \inf\left\{ \sum_{i=1}^\infty x_i^s : \delta\ge x_i \ge 0, \sum_{i=1}^\infty x_i = 1\right\}.$$

For $s=1$, it implies $H^1_\delta([0,1]) = 1$ for all $\delta$, so $H^1([0,1]) = 1$. When $s<1$, $x_i^s \ge x_i$. So

$$H^s_\delta \ge 1$$

for all $\delta$ and so $H^s([0,1]) \neq 0$. When $s>1$, choose the following covering (where $1/n \le \delta$)

$$\{ [0,1/n], [1/n, 2/n], \cdots, [(n-1)/n, 1]\}.$$

Then

$$H^s_\delta([0,1]) \le \sum_{i=1}^n \frac{1}{n^s} = \frac{1}{n^{s-1}} \to 0$$

as $n\to\infty$. This implies

$$H^s_\delta([0,1]) = 0$$

for all $\delta $ and so $H^s([0,1]) = 0$ for all $s>1$. Thus the Hausdorff dimension is one.

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If you are familiar with measure theory then you may note that the restriction to $[0,1]$ of Lebesgue measure $(\lambda)$ is $1$-Ahlfors regular, i.e. for $x\in [0,1]$, $0<r<1/2$: $$ r\leq \lambda(B(x,r)) \leq 2r $$ Given any countable open cover you may replace by a cover using intervals (without changing Hausdorff measure) and then by countable sub-additivity: $$ 1\leq \sum_{k\geq 1} \lambda(U_i) \leq \sum_k |U_i|$$ which implies that dimH is at least 1. Taking an equipartition gives an upper bound equal to 1.

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