Definitions: Let $A\subseteq\mathbb{R.}$ Define the diameter of $A$ by $$|A| = \sup_{x,y\in A}|x-y|.$$
For $A\subseteq\mathbb{R}$ with $|A|<\infty$ and $\alpha>0,$ define the $\alpha$-cover length of $A$ by $$H_\alpha(A) = \inf\bigg\{ \sum_{n=1}^\infty |S_n|^\alpha:\{S_n\}_{n\in\mathbb{N}} \text{ is a countable cover of }A \bigg\}.$$ The Haudorff dimension of $A$ is defined by $$\dim_H(A)=\inf \{\alpha >0:H_\alpha(A)=0\}.$$
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In class, we have proven that $$\dim_H(A) = \sup\{\alpha\geq 0:H_\alpha(A)>0\}.$$
Question: To prove $\dim_H(A) = c\in\mathbb{R},$ is it enough to prove that there exists a single countable cover such that $H_\alpha(A)>0$ if $\alpha<c, H_\alpha(A)=0$ if $\alpha>c?$
For example, let us consider the following propostion.
Proposition: Let $C$ be the Cantor Middle Third set. Prove that $$\dim_H(C)=\log_32.$$
To prove the above proposition, one can let $\alpha = \log_32$ and consider $\alpha'>\alpha.$ For $n\geq 1,$ let $I_1,...,I_{2^n}$ be the set of subintervals that made up of $C_n,$ where $$C=\bigcap_{n=1}^\infty C_n.$$ Clearly $\{I_1,...,I_{2^n}\}$ is a cover for $C.$ Note that $$\sum_{k=1}^{2^n}|I_k|^{\alpha'} = \sum_{k=1}^{2^n}(3^{-n})^{\alpha'} = 2^n(3^{-n\alpha'}) = e^{n(\ln 2-\alpha' \ln3)} = e^{n \ln 3(\alpha - \alpha')}.$$ Since $\alpha - \alpha' <0,$ as $n\to\infty,$ the above sum tends to $0.$ So it implies that $$\dim_H(C) \leq \log_32.$$
After this, one needs to prove that $\dim_H(C)\geq \log_32$ also holds.
But by using supremum definition of $\dim_H(C)$ and the cover $\{I_1,...,I_{2^n}\},$ can we conclude that $$\dim_H(C) = \log_32?$$
More precisely, for any $\alpha<\alpha',$ one can use the same cover and calculation above to obtain $$\sum_{k=1}^{2^n}|I_k|^{\alpha'} = e^{n\ln 3(\alpha-\alpha')}.$$ Since $\alpha-\alpha'>0,$ therefore $$\dim_H(C) \geq \log_32.$$
No, a single cover won't do. In order to show that $\dim_H (C) \geq \log_3 2$ using the result proved in class you have to show that $H_{\alpha} (C) >0$ for some $\alpha \geq \log_3 2$ and $H_{\alpha} (C) $ is defined as an infimum over all covers. For this you have to consider an arbitrary cover.