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\def\Bbb{\mathbb}
\def\Haus{{\cal H}}
\def\dist{{\rm{dist}}}
\def\area{{\rm{area}}}
\def\dist{{\rm{dist}}}
\def\diam{{\rm{diam}}}
\def\dim{{\rm{dim}}}
\def\supp{{\rm{supp}}}
\def\Bbb{\mathbb}
\def\reals{\Bbb R}
\def\cal{\mathcal}
\newenvironment{pf}   {\begin{proof}}{\end{proof}}

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\documentclass[12pt]{amsart}  % use this with LaTeX2e 

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% other helpful packages to include here if you want to 
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\usepackage{amssymb,psfig,verbatim}    %use with LaTeX2e

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\theoremstyle{plain}                    %------- 'regular' theorem types
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{example}[thm]{Example}
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%% Introduction %%%%%%%%%%%%%%%%%%%%

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%\section{Introduction}
%\subsection{}

%\begin{enumerate}

{\bf Exercise 1.11:} Consider $[0,1]$ with the metric 
$d(x,y) = \sqrt{|x-y|}$. Show that the interval has 
dimension 2 with this metric.

{\bf Solution 1.11}
The problem does not specify Hausdorff or Minkowski 
dimension, so we will consider both.

First, we give an upper bound for Minkowski dimension.
Cover $I=[0,1]$ by $n$ intervals of the form $[\frac kn, 
\frac {k+1},n]$ for $k=0, \dots , n-1$. Each of this 
has diameter $1/\sqrt{n}$ in the given metric, so 
$N(I,1/\sqrt{n}) = n$. Hence 
$N(I, \epsilon) = n$ if $ n^{-1/2}  \leq \epsilon
\leq (n-1)^{1/2}$. Hence 
$$   \frac 12 \log (n-1) \leq\log 1/\epsilon \leq \frac 12 \log n,$$
and so 
$$ \limsup_{ \epsilon \to \infty} 
     \frac { \log N(I, \epsilon)}{ \log 1/\epsilon} 
 \leq  \limsup_{n\to \infty} 
     \frac { \log n }{  \frac 12\log  n-1} = 2.$$
Therefore the upper Minkowski dimension is at most $2$.

Next, we give a lower bound for the Hausdorff dimension. 
Consider usual Lebesgue measure  $m$ on $I = [0,1]$. A subinterval 
$J= [x,y]$ gets measure $|x-y|$ but has diameter $\sqrt{|x-y|}$ (remember 
the unusual metric). Thus for every interval  $J$, 
$$   m(J)  = \diam(J)^2.$$
By the mass distribution principle, the Hausdorff dimension is 
at least $2$.

Since the lower bound for Hausdorff dimension and the upper bound 
for the upper  Minkowski dimension agree, and the lower Minkowski 
dimension must be trapped between them, all three numbers agree. 
Thus Minkowski dimension exists and equals $2$, which is the same
as the Hausdorff dimension. 

 

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\begin{comment}
    A comment is a way of preventing text from being 
included into the TeX form.  You can read this when 
you edit the file but it is invisible to TeX (and easier 
that putting the percent sign in front of every line.

\end{comment}
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