齐次Moran集是一类重要的分形集,其Hausdorff维数($ \dim_{H}E $),上、下盒维数($ \overline{\dim}_{B}E $与$ \underline{\dim}_{B}E $)与packing维数($ \dim_{P}E $)^[1]被国内外学者广泛的研究,丰德军、文志英和吴军指出了三种维数值的具体范围,并说明了这些维数达到最大值与最小值的具体集合^[2].文志英和吴军^[3]利用对基本区间间隔的要求定义了一类齐次Moran集,称为齐次完全集,并得到其在一定条件下Hausdorff维数的计算表达式,而后王向阳和吴军^[4]又得到了齐次完全集在同样条件下上盒维数和packing维数的计算表达式.胡晓梅^[5]利用基本区间连成的连通分支构造了一类特殊的齐次Moran集,称为$ \{m_{k}\} $ -拟齐次Cantor集并讨论了其上盒维数和packing维数.

本文在齐次Moran集的基础上加入连通分支的概念,并利用连通分支与其间隔构造了一类比齐次完全集更广泛的齐次Moran集,称为$ \{m_{k}\} $ -拟齐次完全集,并得到了关于该集合上盒维数和packing维数的若干结果.

本文在第二节介绍一些预备知识,包括齐次Moran集、齐次完全集与$ \{m_{k}\} $ -拟齐次完全集的定义;第三节给出主要结果;第四节对主要结果进行证明.

设$ \{n_{k}\}_{k\geq 1} $为正整数序列, $ n_{k}\geq 2 $, $ \{c_{k}\}_{k\geq 1 } $为正实数序列, $ 0 < c_{k} < 1 $,且对$ \forall k\geq 1 $有$ n_{k}c_{k} < 1 $.对$ \forall k\geq 1 $,设$ D_{k} = \{\sigma_{1}\sigma_{2}\cdots \sigma_{k}:1\leq \sigma_{j} \leq n_{j}, 1\leq j \leq k\} $,记$ D = \bigcup\limits_{k = 0}^{\infty}D_{k} $,特别地, $ D_{0} = \{\emptyset\} $.若$ \sigma = \sigma_{1}\sigma_{2}\cdots \sigma_{k} \in D_{k} $, $ \tau = \tau_{1}\tau_{2}\cdots \tau_{m}\in D_{m} $,则记$ \sigma\ast \tau = \sigma_{0}\cdots \sigma_{k}\tau_{1}\cdots\tau_{m} \in D_{k+m} $.

定义2.1^[3] (齐次Moran集)  设$ I\subset \mathbb{R} $为非空闭区间,如果$ I $的闭区间族$ {\cal F} = \{I_{\sigma}:\sigma\in D\} $满足:

(ⅰ) $ I_{\emptyset} = I $;

(ⅱ)对$ \forall k\geq 0 $, $ \sigma \in D_{k} $, $ I_{\sigma\ast 1}, I_{\sigma\ast 2}, \cdots, I_{\sigma\ast n_{k+1}} $是$ I_{\sigma} $的从左至右排列的闭子区间,且对$ \forall i\not = j $, $ I_{\sigma \ast i} $的内部与$ I_{\sigma \ast j} $的内部交集为空;

(ⅲ)对$ \forall k\geq 1 $, $ \sigma\in D_{k-1} $, $ 1\leq j\leq n_{k} $,有$ \frac{|I_{\sigma\ast j}|}{|I_{\sigma}|} = c_{k} $,其中$ |A| $表示集合$ A $的直径.

则称$ \mathcal {F} $具有齐次Moran结构.记$ E = E({\cal F}): = \bigcap\limits_{k\geq 1} \bigcup\limits_{\sigma\in D_{k}}I_{\sigma} $,称$ E $为满足$ (I, \{n_{k}\}, \{c_{k}\}) $的齐次Moran集,并用$ {\cal M}(I, \{n_{k}\}, \{c_{k}\}) $表示所有由$ I $, $ \{n_{k}\}_{k\geq 1} $, $ \{c_{k}\}_{k\geq 1 } $决定的齐次Moran集集族.

令$ {\cal F}_{k} = \{I_{\sigma}:\sigma\in D_{k}\} $,则$ {\cal F} = \bigcup\limits_{k\geq 0}{\cal F}_{k} $.称$ {\cal F}_{k} $中的元素为$ E $的$ k $阶基本区间.

齐次Moran集是一种非常重要的Moran集^[2],其结构与性质被国内外学者广泛研究,关于Moran集与齐次Moran集维数的更多经典结论参见文献[6-12].

对齐次Moran集基本区间之间的间隔做一定的要求,可以定义一类特殊的齐次Moran集:齐次完全集.

定义2.2^[3] (齐次完全集)  设$ E\in {\cal M}(I, \{n_{k}\}, \{c_{k}\}) $,若$ E $满足:存在非负实数序列$ \{\eta_{k, j}:k\geq 1, 0 \leq j\leq n_{k}\} $,使得对$ \forall k\geq 0 $, $ \sigma\in D_{k} $, $ 1 \le i \le n_{k+1}-1 $,有$ \min(I_{\sigma\ast 1})-\min(I_{\sigma}) = \eta_{k+1, 0} $, $ \min(I_{\sigma\ast (i+1)})-\max(I_{\sigma\ast i}) = \eta_{k+1, i} $, $ \max(I_{\sigma})-\max(I_{\sigma\ast n_{k+1}}) = \eta_{k+1, n_{k+1}} $.则称$ E = E(I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, j}\}): = \bigcap\limits_{k\geq 1}\bigcup\limits_{\sigma\in D_{k}}I_{\sigma} $为满足$ (I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, j}\}) $的齐次完全集,并用$ {\cal J}(I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, j}\}) $表示所有由$ I $, $ \{n_{k}\}_{k\geq 1} $, $ \{c_{k}\}_{k\geq 1} $, $ \{\eta_{k, j}\} $决定的齐次完全集集族.

注2.1  华苏在文献[6]中定义了一类著名的特殊Moran集,称为广义自相似集.由定义$ 2.2 $易得齐次完全集就是一维齐次情形下的广义自相似集.

我们在齐次Moran集的基础上利用由基本区间构成的连通分支与其间隔构造了一类比齐次完全集更广泛的齐次Moran集,称为$ \{m_{k}\} $ -拟齐次完全集.

定义2.3 ($ \{m_{k}\} $ -拟齐次完全集)  设$ E \in {\cal M}(I, \{n_{k}\}, \{c_{k}\}) $,设$ I_{\sigma} $是$ E $的$ k-1 $阶基本区间, $ \sigma\in D_{k-1} $,且其$ k $阶基本区间$ I_{\sigma\ast 1} $, $ I_{\sigma\ast 2} $, $ \cdots $, $ I_{\sigma\ast n_{k}} $任意相连形成$ m_{k}(1\le m_{k} \le n_{k}) $个连通分支,称为$ k $阶连通分支,记为$ J_{\sigma\ast 1} $, $ J_{\sigma\ast 2} $, $ \cdots $, $ J_{\sigma\ast m_{k}} $.若$ E $满足:

(ⅰ) $ J_{\emptyset} = J = I $;

(ⅱ)存在非负实数序列$ \{\eta_{k, l}^{*}\} $, $ k\geq 1 $, $ 0\leq l\leq m_{k} $,满足$ \min_{1\le l \le m_{k}-1}{\eta^{*}_{k, l}} > 0 $,且对$ \forall k\geq0 $, $ \sigma\in D_{k} $,有

$ \min(J_{\sigma\ast 1})-\min(I_{\sigma}) = \eta^{*}_{k+1, 0}, \ \min(J_{\sigma\ast (i+1)})-\max(J_{\sigma\ast i}) = \eta^{*}_{k+1, i}\ (1\leq i\leq m_{k+1}-1), $

$ \max(I_{\sigma})-\max(J_{\sigma\ast m_{k+1}}) = \eta^{*}_{k+1, m_{k+1}}. $

则称$ E = E(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}): = \bigcap\limits_{k\geq 1}\bigcup\limits_{\sigma\in D_{k}}I_{\sigma} $为满足$ (I, J, \{n_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $的$ \{m_{k}\} $ -拟齐次完全集,并用$ {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $表示所有由$ I $, $ J $, $ \{n_{k}\} $, $ \{c_{k}\}_{k\geq 1 } $, $ \{\eta^{*}_{k, l}\} $决定的$ \{m_{k}\} $ -拟齐次完全集集族.

注2.2  定义$ 2.3 $中,若$ m_{k} = n_{k} $,则$ E $就是齐次完全集, $ E\in {\cal J}(I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, l}\}) $,其中$ \eta_{k, l} = \eta^{*}_{k, l} $对任意$ k\geq 1 $, $ 0\leq l\leq m_{k} $成立；若$ m_{k} = n_{k} $, $ \eta^{*}_{k, 0} = \eta^{*}_{k, m_{k}} = 0 $,且$ \eta^{*}_{k, 1} = \cdot\cdot\cdot = \eta^{*}_{k, m_{k}-1} $对任意$ k\geq 1 $, $ 1\leq l\leq m_{k}-1 $成立,则$ E $就是齐次Cantor集^[2].

设$ s^{*} = \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}}{-\log c_{1}c_{2}\cdots c_{k}} $, $ t^{*} = \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k+1}}{-\log c_{1}c_{2}\cdots c_{k}+\log n_{k+1}} $,本文的主要结果如下.

定理3.1  设$ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $,若$ \sup_{k}\{m_{k}\} < \infty $,则

(3.1) $ \begin{eqnarray} \overline{\dim}_{B}E = \dim_{P}E = s^{*}. \end{eqnarray} $

例3.1  设$ E $为$ \{m_{k}\} $ -拟齐次Cantor集(参见文献[5]),则$ E $一定为$ \{m_{k}\} $ -拟齐次完全集,其中对$ \forall k\geq1 $,都有$ \eta^{*}_{k, 0} = \eta^{*}_{k, m_{k}} = 0 $,且$ \eta^{*}_{k, 1} = \cdot\cdot\cdot = \eta^{*}_{k, m_{k}-1} $,由定理3.1,若$ \sup_{k}\{m_{k}\} < \infty $,则$ \overline{\dim}_{B}E = \dim_{P}E = s^{*} $,从而定理3.1推广了文献[5,定理2.3]的的结论.

定理3.2  设$ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $,若对$ \forall k\geq1 $,存在正常数$ \tilde{c_{1}} $, $ \tilde{c_{2}} $, $ \tilde{c_{3}} $, $ \tilde{c_{4}} $, $ \alpha(\alpha \ge 1) $,使得$ E $满足以下三个条件之一:

(A) $ \max\limits_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l}\leq \tilde{c_{1}} \cdot c_{1}c_{2}\cdots c_{k} $,且$ \frac{n_{k}}{m_{k}}\leq \alpha $;

(B) $ m_{k}\cdot \min\limits_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l} \geq \tilde{c_{2}} \cdot c_{1}c_{2}\cdots c_{k-1} $;

(C) $ \max\limits_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l}\leq \tilde{c_{3}} \cdot \min\limits_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l} $,且$ \frac{n_{k}}{m_{k}}\leq \alpha $.

则

(3.2) $ \begin{equation} \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log \bigg (\frac{\sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+n_{k+1}c_{1}c_{2}\cdots c_{k+1}}{m_{k+1}}\bigg)}\le\overline{\dim}_{B}E\leq \limsup\limits_{k\rightarrow \infty} \frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}}. \end{equation} $

推论3.1  设$ E $是满足定理3.2条件的$ \{m_{k}\} $ -拟齐次完全集,且对$ \forall k\geq1 $,都有$ \eta^{*}_{k, 0} = \eta^{*}_{k, m_{k}} = 0 $,则

(3.3) $ \begin{equation} \overline{\dim}_{B}E = \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}} {-\log\bigg(\frac{\sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+n_{k+1}c_{1}c_{2}\cdots c_{k+1}}{m_{k+1}}\bigg)} = \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}}. \end{equation} $

注3.1   $ \eta^{*}_{k, 0} = \eta^{*}_{k, m_{k}} = 0 $意味着对$ \forall\sigma\in D_{k} $, $ k\geq 0 $, $ I_{\sigma} $与$ I_{\sigma\ast 1} $的左端点相同,且与$ I_{\sigma\ast n_{k}} $的右端点相同.

例3.2  设$ E $为齐次完全集, $ E\in {\cal J}(I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, l}\}) $$ (l $满足$ 0\le l\le {n_{k}}) $,则$ E $一定为$ \{m_{k}\} $ -拟齐次完全集, $ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $ (其中$ 1\le m_{k} \le n_{k} $且$ l $满足$ 0\le l\le {m_{k}}) $,若$ E $满足对$ \forall k\geq1 $,都有$ \eta_{k, 0} = \eta_{k, n_{k}} = 0 $,则对$ \forall k\geq1 $,都有$ \eta^{*}_{k, 0} = \eta^{*}_{k, m_{k}} = 0 $.从而有:若$ E\in {\cal J}(I, \{n_{k}\}, \{c_{k}\}, \{\eta_{k, l}\}) $,且对$ \forall k\geq1 $,都有$ \eta_{k, 0} = \eta_{k, n_{k}} = 0 $,则

(ⅰ)若$ E $满足文献[4]中定理1.4的条件(A),且存在正常数$ \alpha \ge 1 $,使得$ \frac{n_{k}}{m_{k}}\leq \alpha $对$ \forall k\geq1 $成立,或$ E $满足文献[4]中定理1.4的(B)与(C)中任何一个条件(若满足(B)或(C),则一定有$ \frac{n_{k}}{m_{k}} = 1 $$) $,可验证$ E $为满足定理3.2和推论3.1的条件的$ \{m_{k}\} $ -拟齐次完全集,从而$ (3.3) $式对$ E $成立,注意到此时有$ m_{k}\le n_{k} \le \alpha m_{k} $,从而$ (3.3) $式中的$ m_{k+1} $可用$ n_{k+1} $代替,即$ E $满足文献[4]中$ (1.4) $式的上盒维数公式;

(ⅱ)若$ E $满足文献[4,定理1.4]中条件(D),即$ \sup_{k}\{n_{k}\} < \infty $,则$ E $为满足定理3.1条件的$ \{m_{k}\} $ -拟齐次完全集(由$ \sup_{k}\{n_{k}\} < \infty $知$ \sup_{k}\{m_{k}\} < \infty) $,从而$ (3.1) $式对$ E $成立,又由$ \sup_{k}\{n_{k}\} < \infty $知$ s^{*} = t^{*} $,从而$ \overline{\dim}_{B}E = \dim_{P}E = t^{*} $,即文献[4]的$ (1.4) $式成立(由$ \eta_{k, 0} = \eta_{k, n_{k}} = 0 $立得).

从而定理3.1与推论3.1推广了文献[4,定理1.4]在附加条件$ \eta_{k, 0} = \eta_{k, n_{k}} = 0 $与$ \frac{n_{k}}{m_{k}}\leq \alpha $$ (\forall k\geq1) $下(若满足条件(B)-(D)中的任何一个条件则只需附加条件$ \eta_{k, 0} = \eta_{k, n_{k}} = 0) $上盒维数的结论.

定理3.1和3.2的证明需要用到下面的2个引理.

引理4.1^[2]  设齐次Moran集$ E\in{\cal M}(I, \{n_{k}\}, \{c_{k}\}) $,则有

$ s^{*} \leq \dim_{p}E\leq \overline{\dim}_{B}E\leq t^{*}, $

引理4.2^[13]  设$ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $,则

$ \overline{\dim}_{B}E = \inf \bigg\{s>0:\sum\limits_{k = 1}^{\infty}n_{1}n_{2}\cdots n_{k}\sum\limits_{i = 0}^{m_{k+1}}(\eta^{\ast}_{k+1, i})^{s}<\infty\bigg\}. $

由引理4.1, $ \dim_{p}E \geq s^{*} $,所以只需证明$ \overline{\dim}_{B}E\leq s^{*} $.若$ s^{*} = 1 $则显然成立,下设$ s^{*} < 1 $,则对$ \forall 1 > t > s^{*} = \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}}{-\log c_{1}c_{2}\cdots c_{k}} $, $ \exists\varepsilon > 0 $, $ k_{0} > 0 $,使得对$ \forall k\geq k_{0} $,有

(4.1) $ \begin{eqnarray} t>\frac{\log n_{1}n_{2}\cdots n_{k}}{-\log c_{1}c_{2}\cdots c_{k}}+\varepsilon. \end{eqnarray} $

令$ b_{k} = t-\frac{\log n_{1}n_{2}\cdots n_{k}}{-\log c_{1}c_{2}\cdots c_{k}} > \varepsilon $,可得$ n_{1}n_{2}\cdots n_{k}(\delta_{k})^{t} = (\delta_{k})^{b_{k}} $.由$ n_{k} $, $ c_{k} $定义,得$ \delta_{k} = c_{1}c_{2}\cdots c_{k}\leq2^{-k} $,又由$ m_{k}\geq1 $,得到$ m_{k}+1\ge 2 $且$ m_{k}+1\le 2m_{k} $对任意$ k\ge 1 $成立.由函数$ x^{s}(0 < s < 1) $的凹凸性,有

(4.2) $ \begin{eqnarray} \sum\limits_{i = 0}^{m_{k+1}}(\eta_{k+1, i}^\ast)^{t} \leq(m_{k+1}+1) \bigg(\frac{\sum\limits_{i = 0}^{m_{k+1}}\eta_{k+1, i}^\ast}{m_{k+1}+1}\bigg)^{t} = (m_{k+1}+1) \bigg(\frac{(1-c_{k+1}n_{k+1})\delta_{k}}{m_{k+1}+1}\bigg)^{t}. \end{eqnarray} $

记$ m = \sup_{k}\{m_{k}\} < \infty $,从而根据(4.1)式和(4.2)式有

$ \begin{eqnarray*} \sum\limits_{k = k_{0}}^{\infty}n_{1}n_{2}\cdots n_{k}\sum\limits_{i = 0}^{m_{k+1}}(\eta^{\ast}_{k+1, i})^{t} &\leq &\sum\limits_{k = k_{0}}^{\infty}n_{1}n_{2}\cdots n_{k}(m_{k+1}+1) \bigg (\frac{(1-c_{k+1}n_{k+1})\delta_{k}}{m_{k+1}+1}\bigg)^{t}\\ &\leq&2\sum\limits_{k = k_{0}}^{\infty}n_{1}n_{2}\cdots n_{k}m_{k+1} \bigg(\frac{\delta_{k}}{m_{k+1}+1}\bigg)^{t}\\ &\leq&2m\sum\limits_{k = k_{0}}^{\infty}n_{1}n_{2}\cdots n_{k}\bigg(\frac{\delta_{k}}{m_{k+1}+1}\bigg)^{t}\\ & = &2m\sum\limits_{k = k_{0}}^{\infty}n_{1}n_{2}\cdots n_{k}(\delta_{k}^{t})(m_{k+1}+1)^{-t}\\ & = &2m\sum\limits_{k = k_{0}}^{\infty}(\delta_{k}^{b_{k}})(m_{k+1}+1)^{-t} \leq2m\sum\limits_{k = k_{0}}^{\infty}(\delta_{k}^{b_{k}})2^{-t}\\ &\leq&2m\sum\limits_{k = k_{0}}^{\infty}2^{-k\varepsilon}\cdot 1<\infty, \end{eqnarray*} $

从而根据引理4.2,有$ \overline{\dim}_{B}E\leq t $,由$ t $的任意性可得$ \overline{\dim}_{B}E\leq s^{*} $,从而定理3.1得证.

先估计$ \overline{\dim}_{B}E $的上界,有如下命题.

命题4.1  设$ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $,则有

$ \overline{\dim}_{B}E\leq \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}}. $

证  对$ \forall 1 > l > \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}} $, $ \exists\varepsilon > 0 $, $ k_{1} > 0 $,使得对$ \forall k\geq k_{1} $,有

$ l> \frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}}+\varepsilon, $

从而

$ (\varepsilon-l)\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}>\log n_{1}n_{2}\cdots n_{k}m_{k+1}, $

即

(4.3) $ \begin{eqnarray} n_{1}n_{2}\cdots n_{k}m_{k+1}<\bigg (\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}\bigg)^{\varepsilon-l}. \end{eqnarray} $

仍记$ \delta_{k} = c_{1}c_{2}\cdots c_{k}\le 2^{-k} $, $ \forall k\geq0 $,由函数$ x^{s}(0 < s < 1) $的凹凸性,有

(4.4) $ \begin{eqnarray} \sum\limits_{i = 0}^{m_{k+1}}(\eta_{k+1, i}^\ast)^{l} \leq(m_{k+1}+1)\bigg(\frac{\sum\limits_{i = 0}^{m_{k+1}}\eta_{k+1, i}^\ast}{m_{k+1}+1}\bigg)^{l} = (m_{k+1}+1)\bigg(\frac{(1-c_{k+1}n_{k+1})\delta_{k}}{m_{k+1}+1}\bigg)^{l}. \end{eqnarray} $

由$ m_{k}\ge 1 $知$ m_{k}+1\le 2m_{k} $对任意$ k\ge1 $成立,再由(4.3)式和(4.4)式得

$ \begin{eqnarray*} \sum\limits_{k = k_{1}}^{\infty}n_{1}n_{2}\cdots n_{k}\sum\limits_{i = 0}^{m_{k+1}}(\eta^{\ast}_{k+1, i})^{l} &\leq& \sum\limits_{k = k_{1}}^{\infty}n_{1}n_{2}\cdots n_{k}(m_{k+1}+1) \bigg (\frac{(1-c_{k+1}m_{k+1})\delta_{k}}{m_{k+1}+1}\bigg)^{l}\\ &\leq&2\sum\limits_{k = k_{1}}^{\infty}n_{1}n_{2}\cdots n_{k}m_{k+1} \bigg(\frac{\delta_{k}}{m_{k+1}+1}\bigg)^{l}\\ &\leq&2\sum\limits_{k = k_{1}}^{\infty}\bigg(\frac{\delta_{k}}{m_{k+1}}\bigg)^{\varepsilon-l} \bigg(\frac{\delta_{k}}{m_{k+1}}\bigg)^{l}\\ & = &2\sum\limits_{k = k_{1}}^{\infty}\bigg(\frac{\delta_{k}}{m_{k+1}}\bigg)^{\varepsilon} \leq2\sum\limits_{k = k_{1}}^{\infty}2^{-k\varepsilon}<\infty. \end{eqnarray*} $

由引理4.2得$ \overline{\dim}_{B}E\leq l $.根据$ l $的任意性可知$ \overline{\dim}_{B}E\leq \limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log\frac{c_{1}c_{2}\cdots c_{k}}{m_{k+1}}} $.

下面估计$ \overline{\dim}_{B}E $的下界.对$ \forall k\geq1 $,定义$ L_{k} = \eta^{\ast}_{k, 0} $, $ R_{k} = \eta^{\ast}_{k, n_{k}} $, $ \overline{\alpha}_{k} = \max_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l} $, $ \underline {\alpha}_{k} = \min_{1 \leq l\leq m_{k}-1} \eta^{\ast}_{k, l} $.则对$ \forall k\geq1 $,有

$ \frac{\sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+m_{k+1}c_{1}c_{2} \cdots c_{k+1}}{m_{k+1}} = \frac{c_{1}c_{2}\cdots c_{k+1}-L_{k+1}-R_{k+1}}{m_{k+1}}. $

命题4.2  设$ E \in {\cal J}^{*}(I, J, \{n_{k}\}, \{m_{k}\}, \{c_{k}\}, \{\eta^{*}_{k, l}\}) $,且$ E $满足定理3.2的条件,则

$ \overline{\dim}_{B}E\geq\limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log \bigg (\frac{\sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+n_{k+1}c_{1}c_{2}\cdots c_{k+1}}{m_{k+1}}\bigg)}. $

证  令$ r_{1} = \frac{1-L_{1}-R_{1}}{m_{1}} $,对$ \forall k\geq2 $,令$ r_{k} = \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}} $.固定足够大的$ k $,下面按照定理3.2的三个条件分别估计$ \frac{\log N_{r_{k}}(E)}{-\log r_{k}} $.

条件(A)  此时有$ \overline{\alpha}_{k}\leq \tilde{c_{1}} \cdot c_{1}c_{2}\cdots c_{k} $,且$ \frac{n_{k}}{m_{k}}\leq \alpha $成立.定义$ M_{k} $为与直径为$ r_{k} $的球相交的$ k $阶连通分支的最大个数,则有

$ \begin{eqnarray*} (M_{k}-2)c_{1}c_{2}\cdots c_{k}&\leq &r_{k} = \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}}\\ &\leq&\frac{n_{k}c_{1}c_{2}\cdots c_{k}+(m_{k}-1)\overline{\alpha}_{k}}{m_{k}} \\ & \leq&\frac{n_{k}c_{1}c_{2}\cdots c_{k}+(m_{k}-1)\tilde{c_{1}} \cdot c_{1}c_{2}\cdots c_{k}}{m_{k}}\\ &\leq&\frac{n_{k}c_{1}c_{2}\cdots c_{k}+(n_{k}-1)\tilde{c_{1}} \cdot c_{1}c_{2}\cdots c_{k}}{m_{k}}\\ & \leq&\frac{c_{1}c_{2}\cdots c_{k}(n_{k}+n_{k}\tilde{c_{1}})}{m_{k}}, \end{eqnarray*} $

所以有

$ M_{k}-2\leq \frac{n_{k}+n_{k}\tilde{c_{1}}}{m_{k}}\leq \alpha(1+\tilde{c_{1}}), $

即$ M_{k}\leq 2+\alpha(1+\tilde{c_{1}}) $.从而

(4.5) $ \begin{equation} \frac{\log N_{r_{k}}(E)}{-\log r_{k}}\geq\frac{\log\frac{1}{M_{k}}n_{1}n_{2}\cdots n_{k-1}m_{k}}{-\log r_{k}}\geq \frac{\log n_{1}n_{2}\cdots n_{k-1}m_{k}-\log (2+\alpha(1+\tilde{c_{1}}))}{-\log\frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}}}. \end{equation} $

条件(B)  此时有$ m_{k}\cdot \underline{\alpha_{k}}\geq \tilde{c_{2}} \cdot c_{1}c_{2}\cdots c_{k-1} $,从而

$ \frac{1}{2}\underline{\alpha}_{k}\geq \frac{\tilde{c_{2}}}{2} \cdot \frac{c_{1}c_{2}\cdots c_{k-1}}{m_{k}}\geq \frac{\tilde{c_{2}}}{2} \cdot \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}} = \frac{\tilde{c_{2}}}{2}\cdot r_{k}. $

这意味着每个直径为$ \frac{\tilde{c_{2}}}{2}\cdot r_{k} $的球最多交2个$ k $级连通分支,所以直径为$ r_{k} $的球最多交$ 2(\frac{2}{\widetilde{c_{2}}}+1): = T_{k} $个$ k $级连通分支.从而

(4.6) $ \begin{equation} \frac{\log N_{r_{k}}(E)}{-\log r_{k}}\geq\frac{\log\frac{1}{T_{k}}n_{1}n_{2}\cdots n_{k-1}m_{k}}{-\log r_{k}}\geq \frac{\log n_{1}n_{2}\cdots n_{k-1}m_{k}-\log (2(\frac{2}{\widetilde{c_{2}}}+1))}{\log \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}}}. \end{equation} $

条件(C)  此时有$ \overline{\alpha}_{k}\leq \widetilde{c_{3}}\cdot \underline{\alpha}_{k} $,且$ \frac{n_{k}}{m_{k}}\leq \alpha $成立.分为以下两种情形讨论.

情形1  $ \overline{\alpha}_{k}\leq \widetilde{c_{3}}\cdot \underline{\alpha}_{k} $且$ \underline{\alpha}_{k} < c_{1}c_{2}\cdots c_{k} $,此时有$ \overline{\alpha}_{k}\leq \tilde{c_{3}} \cdot c_{1}c_{2}\cdots c_{k} $,由条件(A)的讨论可得

(4.7) $ \begin{equation} \frac{\log N_{r_{k}}(E)}{-\log r_{k}}\geq\frac{\log\frac{1}{M_{k}}n_{1}n_{2}\cdots n_{k-1}m_{k}}{-\log r_{k}}\geq \frac{\log n_{1}n_{2}\cdots n_{k-1}m_{k}-\log (2+\alpha(1+\tilde{c_{3}}))}{-\log\frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}}}. \end{equation} $

情形2  $ \overline{\alpha}_{k}\leq \widetilde{c_{3}}\cdot \underline{\alpha}_{k} $且$ \underline{\alpha}_{k}\geq c_{1}c_{2}\cdots c_{k} $.则

$ \begin{eqnarray*} c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}&\leq &n_{k}c_{1}c_{2}\cdots c_{k}+(m_{k}-1)\overline{\alpha}_{k}\\ & \leq &n_{k}c_{1}c_{2}\cdots c_{k}+(m_{k}-1)\widetilde{c_{3}}\underline{\alpha}_{k}\\ &\leq& n_{k}c_{1}c_{2}\cdots c_{k}+n_{k}\widetilde{c_{3}}\underline{\alpha}_{k}\\ &\leq& n_{k}\underline{\alpha}_{k}+n_{k}\widetilde{c_{3}}\underline{\alpha}_{k}, \end{eqnarray*} $

可得

$ \underline{\alpha}_{k}\geq \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{n_{k}(1+\widetilde{c_{3}})} \geq \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{\alpha(m_{k}(1+\widetilde{c_{3}}))} = \frac{1}{\alpha(1+\widetilde{c_{3}})}\cdot r_{k}. $

对此条件进行与条件(B)相似的讨论,可以得到

(4.8) $ \begin{equation} \frac{\log N_{r_{k}}(E)}{-\log r_{k}}\geq \frac{\log n_{1}n_{2}\cdots n_{k-1}m_{k}-\log 2(2\alpha(\widetilde{c_{3}}+1)+1)}{\log \frac{c_{1}c_{2}\cdots c_{k-1}-L_{k}-R_{k}}{m_{k}}}. \end{equation} $

注意到当$ k\rightarrow \infty $时,有$ r_{k}\rightarrow 0 $,从而由(4.5)-(4.8)式得到

$ \begin{eqnarray*} \overline{\dim}_{B}E& = &\limsup\limits_{\varepsilon\rightarrow 0} \frac{\log N_{\varepsilon}(E)}{-\log \varepsilon} \geq \limsup\limits_{k\rightarrow \infty} \frac{\log N_{r_{k}}(E)}{-\log r_{k}}\\ & \geq&\limsup\limits_{k\rightarrow \infty}\frac{\log n_{1}n_{2}\cdots n_{k}m_{k+1}}{-\log \bigg(\frac{\sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+n_{k+1}c_{1}c_{2}\cdots c_{k+1}}{m_{k+1}}\bigg)}. \end{eqnarray*} $

命题4.2证毕.

由命题4.1和命题4.2可得定理3.2成立.如果对$ \forall k\geq1 $,都有$ \eta^{*}_{k, 0} = \eta^{*}_{k, , m_{k+1}} = 0 $,则$ c_{1}c_{2}\cdots c_{k} = \sum\limits_{l = 1}^{m_{k+1}-1}\eta^{\ast}_{k+1, l}+n_{k+1}c_{1}c_{2}\cdots c_{k+1} $,从而由定理3.2知推论3.1成立.