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The upper Minkowski dimension of a compact set $A$ in $\mathbb{R}$ is defined as $$ \overline{\dim}_M = \inf \{ \epsilon > 0 : \text{ there is a constant } C(\epsilon) \text{ such that } \mu(A_\delta) \leq C(\epsilon) \delta^{1-\epsilon} \text{ for all } \delta > 0 \} $$

Here $\mu$ is the Lebesgue measure and $$ A_\delta = \{x : \text{dist}(x,A) < \delta \} $$ is the $\delta$-neighborhood of $A$.

How do I compute the upper Minkowski dimension of $$ A = \{ 0,1,\frac{1}{2}, \frac{1}{3}, \ldots \} $$

I see from Wikipedia that the answer should be $\frac{1}{2}$, but I want to know how this is calculated.

I see that $A_\delta$ is the union of the open intervals $(\frac{1}{n} - \delta, \frac{1}{n} + \delta)$ for $n \geq 1$. For any fixed $\delta > 0$, all but finitely many points of $A$ belong to $(-\delta,\delta) \subseteq A_\delta$, so that $\mu(A_\delta)$ is bounded by a constant multiple of $\delta$. However, this constant will depend on $\delta$. I suppose my aim is to show that the constant behaves like $\delta^{-1/2}$ as $\delta$ varies, but I don't see how to do that.

Can someone please help me?

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1 Answer 1

up vote 3 down vote accepted

In practice, I prefer the (equivalent) following definition of upper Minkowski dimension:

$$\overline{\dim}(E) = \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log(1/\delta)}. $$

$N(E, \delta)$ is the number of $\delta$-balls needed to cover the set $E$. (We need $E$ to be compact).

(Throughout, we write $X \approx Y$ if there is a constant $c\geq 1$ such that $\frac{1}{c}X \leq Y \leq c X$. We also write $X \sim Y$, if $\lim (X/Y) = 1$ with respect to some common paramater of $X$ and $Y$.)


EDIT: To see that these notions are equivalent, just notice that $\mu(E_{\delta}) \approx \delta^n N(E,\delta)$. Formally, for a compact set $E \subset \mathbb{R}^d$, we have

$$\begin{align} \overline{\dim}_M(E) &= \inf\left\{\alpha \in [0,d] : \delta^{-n}\mu(E_{\delta}) \lesssim \delta^{-\alpha}\right\} \\ &= \inf \left\{\alpha \in [0,d] : \limsup_{\delta \to 0} N(E,\delta) \cdot \delta^{\alpha} = 0\right\} \\ &= \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log (1/\delta)} \end{align}$$


With this in mind, let's consider how many $\delta$-balls are necessary to cover $E = \{0,1, 1/2, \dots\}$. First, break up the set $E$ into two pieces:

$$\left\{1, \frac{1}{2}, \dots , \frac{1}{n}\right\} \cup \left( \left\{\frac{1}{n+1}, \dots \right\} \cup \{ 0 \} \right) = A \cup B$$

Well, we need $n$ $\delta$-balls to cover $A$ (There are $n$ distinct points), and $\approx n$ $\delta$-balls to cover $B$ (since $| \frac{1}{n+1} - 0| \cdot n \approx 1$). (We needn't be exact with the precise value of $N(E, \delta)$. We really only need an asymptotic value).

Therefore, $N(E, \delta) = n + n \approx n$. However, what is $n$ in terms of $\delta$? As long as $\delta \ll \frac{1}{n} - \frac{1}{n+1} \approx \frac{1}{n^2}$, we are not overlapping. Therefore, we can choose $\delta \approx \frac{1}{n^2}$ or $n \approx 1/\sqrt{\delta}$.

Finally:

$$ \overline{\dim(E)} = \limsup_{\delta \to 0} \frac{\log(1/\sqrt{\delta})}{\log(1 /\delta)} = \frac{1}{2}. $$

This argument can be made as formal as you like, though we really only need an asymptotic value for $N(E, \delta)$, and $\delta$ (since $\log cx \sim \log x$).

The trick with calculating the Minkowski dimension of a set is to let $\delta \to 0$ and $n \to \infty$ in a controlled way. Here, setting $\delta \approx \frac{1}{n^2}$ did the trick.

I would try adapting the argument to find the (upper) Minkwoski dimension of the set $E_{\alpha} = \{1, \frac{1}{2^{\alpha}}, \frac{1}{3^{\alpha}}, \dots \} \cup \{ 0 \}$ (here, $\alpha > 0$). I find these examples really fun to play with! (This also let's one construct a set with Minkowski dimension $\beta$ for any $0 < \beta < 1$).

Furthermore, if you are ever in need of a good textbook or reference on harmonic analysis and Hausdorff or Minkowski dimensions, I would strongly recommend Tom Wolff's "Lectures on harmonic analysis" and Pertti Mattila's "Geometry of Sets and Measures in Euclidean Spaces"(which is basically the gold standard for geometric measure theory!).

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Thank you for your answer and the reference to Wolff's book. I used the general strategy in your answer to prove $\overline{\dim}_M = \frac{1}{2}$ with my original definition. I also worked out how to get the dimension of $E_\alpha$ as you suggested. However, I wasn't able to prove the equivalence of the two definitions. Would you mind indicating how this is proved or giving a reference? –  admchrch Mar 28 '11 at 8:13
    
Another great reference to have (This is the must have book for geometric measure theory) is Pertti Mattila's "Geometry of Sets and Measures in Euclidean Spaces". It is fantastically written, and is quite digestible. –  JavaMan Mar 28 '11 at 16:56

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