Main Theme of the Question: I read (and proved) that the upper box dimension is finitely stable. I think we can mimic the same proof for the lower box dimension, but it is explicitly stated (in Falconer's Fractal Geometry) that the lower box dimension is not finitely stable. I write this post to seek help in finding the flaw in my proof. For a brief word on notation, please skip to the end of the post. If you're interested, please continue reading.
Proof of Finite Stability of Upper Box Dimension:
$\overline\dim_B F$ is finitely stable, i.e. $\overline \dim_B (E\cup F) = \max\{\overline\dim_B E, \overline\dim_B F\}$. To see this, we prove $\le$ and $\ge$ inequalities.
- Since $E\subset E\cup F$ and $F\subset E\cup F$, it is clear that $\overline\dim_B E \le \overline \dim_B (E\cup F)$ and $\overline\dim_B E \le \overline \dim_B (E\cup F)$, hence $\max\{\overline\dim_B E, \overline\dim_B F\} \le \overline \dim_B (E\cup F)$.
- Note that $N_\delta(E\cup F) \le N_\delta(E) + N_\delta(F) \le 2\max\{N_\delta(E), N_\delta(F)\}$. Taking logarithm on both sides, interchanging $\max$ and $\log$ on RHS (because $\log$ is a strictly increasing function), and dividing throughout by $-\log\delta$, we get
$$\frac{\log N_\delta(E\cup F)}{-\log\delta} \le \frac{\log 2}{-\log\delta} + \max\left\{\frac{\log N_\delta(E)}{-\log\delta},\frac{\log N_\delta(F)}{-\log\delta} \right\}$$
Taking $\delta\to 0$, we get $\overline\dim_B(E\cup F)\le \max\{\overline\dim_B E, \overline\dim_B F\}$.
Question: Why can't we repeat exactly the same proof for $\underline\dim_B F$, taking $\underline\lim_{\delta\to 0}$ in place of $\overline\lim_{\delta\to 0}$ everywhere? Am I missing something subtle, that disallows us from taking $\underline\lim_{\delta\to 0}$ in one (or more) of the inequalities? I'm not able to see this, and I'd appreciate any help.
Since $E\subset E\cup F$ and $F\subset E\cup F$, $\underline\dim_B E \le \underline \dim_B (E\cup F)$ and $\underline\dim_B E \le \underline \dim_B (E\cup F)$ still hold. Hence $\max\{\underline\dim_B E, \underline\dim_B F\} \le \underline \dim_B (E\cup F)$ must also continue to hold. My sense is that something is going wrong while taking limits (or maybe somewhere else) in the other direction of the inequality, i.e. $\underline\dim_B(E\cup F)\le \max\{\underline\dim_B E, \underline\dim_B F\}$ must not hold.
Notation:
- $\overline\dim_B F$, $\underline\dim_B F$ and $\dim_B F$:
The lower and upper box-counting dimensions of a subset $F$ of $ℝ^n$ are given by
$$\color{blue}{\underline{\dim}_B F = \underline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}}$$
$$\color{blue}{\overline{\dim}_B F = \overline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}}$$
and the box-counting dimension of $F$ by
$$\color{blue}{\dim_B F = \lim_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}}$$
(if this limit exists), where $N_\delta(F)$ is the smallest number of sets of diameter at most $𝛿$ that cover $F$.
- What is $\underline{\lim}_{\delta\to 0}$ and $\overline{\lim}_{\delta\to 0}$?
$$\color{blue}{\underline{\lim}_{x\to 0} f(x) = \lim_{r\to 0} \inf\{f(x) : 0 < x < r\}}$$
$$\color{blue}{\overline{\lim}_{x\to 0} f(x) = \lim_{r\to 0} \sup\{f(x) : 0 < x < r\}}$$
Related questions on this site: (these do not answer my question – my question is very specific to understanding the flaw in my proposed proof, which mimics the corresponding proof for upper box dimension).
(a) How can I find an example to illustrate that the lower box dimension may not be finitely stable?
(b) Lower Box Dimension Inequality [duplicate]
(c) Simple counter example to show box-dimension is not finitely stable? [duplicate]
References:
Fractal Geometry, Kenneth Falconer.
Thank you!
Best Answer
For the sake of completeness, I am adding an answer based on @XanderHenderson's comments. In general, if $E$ and $F$ are two subsets of $\mathbb R^n$ (or whatever space), and $E\subset F$, then $$\sup(E) \le \sup(F)$$ and $$\inf(E) \ge \inf(F)$$ The upper box dimension is finitely stable since taking supremum on both sides preserves the inequality, while taking infimum reverses it (hence the lower box dimension is not finitely stable).