Can some one explain this property & the proof for me ? It goes like this:

If $F_1,F_2,....$ is a countable sequence of sets then $\dim_H \cup_{i=1}^{\infty}F_i=\sup_{1 \leq i < \infty} \{\dim_HF_i\}$. Since $\dim_H \cup_{i=1}^{\infty}F_i \geq \sup_{1 \leq i < \infty} \{\dim_HF_i\}$ from the monotonicity. For the other direction, if $s>\dim_HF_i$ for all $i$, then $H^s(F_i)=0$, so we have $H^s(\cup_{i=1}^{\infty}F_i)=0$. I know the first direction, but what about the second direction ?


Let $(X_j)_{j \geq 0}$ be a family of subsets of $X$.

If $(U_i^j)_{i \in I}$ is a $\delta$-cover of $X_j$, then $(U_i^j)_{(i,j) \in I \times \mathbb{N}}$ is a $\delta$-cover of $\bigcup\limits_{j \geq 0} X_j$. Thus, $$H_{\delta}^s \left( \bigcup\limits_{j \geq 0} X_j \right) \leq \sum\limits_{j \geq 0} \sum\limits_{i \in I} |U_i^j|^s;$$ taking the minimum over the $\delta$-cover, we obtain $$H_{\delta}^s \left( \bigcup\limits_{j \geq 0}X_j \right) \leq \sum\limits_{j \geq 0} H_{\delta}^s(X_j);$$ when $\delta \to 0$, we get $\displaystyle H^s \left( \bigcup\limits_{j \geq 0} X_j \right) \leq \sum\limits_{j \geq 0} H^s(X_j)$.

Therefore, if for all $j \geq 0$, $H^s(X_j)=0$, we necessarily have $H^s \left( \bigcup\limits_{j \geq 0} X_j \right)=0$. We deduce that $\dim_H \left( \bigcup\limits_{j \geq 0} X_j \right) \leq \sup\limits_{j \geq 0} \dim_H(X_j)$.

  • $\begingroup$ I understand the idea of this proof but one step is unclear to me: why can you exchange sum and limes on the right side of the inequality? What I mean is why can you do this: $\lim_{\delta \to 0} \sum_{j} H_{\delta}^s(X_j)=\sum_{j} \lim_{\delta \to 0} H_{\delta}^s(X_j)$ $\endgroup$ – thehardyreader Dec 2 '17 at 14:49

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