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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 1\\ Bryden Cais}
\medskip
\item Suppose that $X=\{\alpha\in S: f(\alpha)>0\}$ is uncountable and define $X_n=\{\alpha\in S: f(\alpha)>1/n\}$.
Then clearly
$$X=\bigcup_{n=1}^{\infty} X_n.$$
If each $X_n$ were finite, then $X$ would be a countable union of finite sets and hence countable.  It follows
that there exists $n_0$ such that $X_{n_0}$ is infinite.  Let $T_r$ be any finite subset of $X_{n_0}$ consisting of
$r$ elements (such a subset exists for each $r>0$ since $X_{n_0}$ is infinite).  We then have
$$\sum_{\alpha\in S}f(\alpha)\geq \sum_{\alpha\in X_{n_0}}f(\alpha)\geq  \sum_{\alpha\in T_r}\frac{1}{n}=\frac{r}{n}$$
for every positive integer $r$.  It follows that $\sum_{\alpha\in S}f(\alpha)\not<\infty$, which is the contrapositive.

\item Let $x,y\in E$.  Without loss of generality, we may assume that $x<y$. Suppose that $[x,y]\nsubseteq E$.
Then there exists some $z\in \R\setminus E$ with $x<z<y$.
Observe that since
$$a(y)=\sup \{u\in \R\setminus E:u<y\}$$ we have $a(y)\geq z.$
Similarly, since
$$b(x)=\inf \{u\in \R\setminus E:x <u\}$$ we have $b(x)\leq z.$
Thus, $b(x)\leq a(y)$
Trivially, we have $a(x)<b(x)$ and $a(y)<b(y)$ so that $(a(x),b(x))\cap (a(y),b(y))=\emptyset.$
Therefore, if $(a(x),b(x))\cap (a(y),b(y))\neq\emptyset$ then $[x,y]\subseteq E$.
It follows that
$$\{u\in \R\setminus E: u < y\}=\{u\in \R\setminus E: u < x\}$$ and similarly that
$$\{u\in \R\setminus E: x < u\}=\{u\in \R\setminus E: y < u\},$$ for otherwise, in any case, we would obtain
some $z\in \R\setminus E$ with $x<z<y$, a contradiction.
This clearly implies that $a(x)=a(y)$ and $b(x)=b(y)$.


\item Let $f:\mathcal{E}\rightarrow \mathcal{C}$ and $g:\mathcal{E}\rightarrow [0,1]$ be given by
\begin{align*}
f(\{\epsilon_n\})&=\sum_{n=0}^{\infty}\frac{2\epsilon_n}{3^n}\\
g(\{\epsilon_n\})&=\sum_{n=0}^{\infty}\frac{\epsilon_n}{2^n}.
\end{align*}
For $\{\epsilon_n\},\{\tilde{\epsilon}_n\}\in \mathcal{E},$ set $m=\min\{n:\epsilon_n\neq\tilde{\epsilon}_n\}.$
\begin{enumerate}
    \item We have
    $$f(\{\epsilon_n\})-f(\{\tilde{\epsilon}_n\})=\sum_{n=0}^{\infty}\frac{2(\epsilon_n-\tilde{\epsilon}_n)}{3^n}=\sum_{n=m}^{\infty}\frac{2(\epsilon_n-\tilde{\epsilon}_n)}{3^n}.$$
    Since $\epsilon_n,\tilde{\epsilon}_n\in \{0,1\}$ we have $|\epsilon_n-\tilde{\epsilon}_n|\leq 1$ so that
    $$\left|\sum_{n=m}^{\infty}\frac{2(\epsilon_n-\tilde{\epsilon}_n)}{3^n}\right|\leq \sum_{n=m}^{\infty}\frac{2}{3^n}=\frac{1}{3^{m-1}}.$$
    On the other hand, the triangle inequality gives
    \begin{align*}
        \left|\sum_{n=m}^{\infty}\frac{2(\epsilon_n-\tilde{\epsilon}_n)}{3^n}\right|\geq \frac{2}{3^m}-\left|\sum_{n=m+1}^{\infty}\frac{2(\epsilon_n-\tilde{\epsilon}_n)}{3^n}\right|,
    \end{align*}
    and as we have just seen, the second term above is at most $1/3^m$.  It follows that we have the bounds
    $$\frac{1}{3^m}\leq |f(\{\epsilon_n\})-f(\{\tilde{\epsilon}_n\})|\leq \frac{1}{3^{m-1}}$$
    for any two {\em different} sequences $\epsilon_n,\tilde{\epsilon}_n$.  Moreover, these bounds are sharp as the
    sequences
    \begin{align*}
    \epsilon_n&=\begin{cases}1 & \quad\text{if}\ n=m\\ 0 & \quad\text{otherwise}\end{cases} &
    \tilde{\epsilon}_n&=\begin{cases}\phantom{-}0 & \quad\text{if}\ n\leq m\\ -1 & \quad\text{otherwise}\end{cases}
    \end{align*}
    and
    \begin{align*}
    \epsilon_n&=\begin{cases}1 & \quad\text{if}\ n\geq m\\ 0 & \quad\text{otherwise}\end{cases} &
    \tilde{\epsilon}_n&\equiv 0
    \end{align*}
    attain the lower and upper bounds, respectively.

    \item Observe that we obviously have $|f(\{\epsilon_n\})-f(\{\tilde{\epsilon}_n\})|\geq 0$ and that
    $$\sum_{n=m+1}^{\infty}\frac{1}{2^n}=\frac{1}{2^m},$$ so that this bound is sharp.  By the same estimates in part (a),
    we find
    $$\left|\sum_{n=m}^{\infty}\frac{(\epsilon_n-\tilde{\epsilon}_n)}{2^n}\right|\leq \sum_{n=m}^{\infty}\frac{1}{2^n}=\frac{1}{2^{m-1}},$$
    so that we have the bounds
    $$0\leq |g(\{\epsilon_n\})-g(\{\tilde{\epsilon}_n\})|\leq \frac{1}{2^{m-1}}.$$
    Moreover, the upper bound is sharp also, as it is attained by the sequences
    \begin{align*}
    \epsilon_n&=\begin{cases}1 & \quad\text{if}\ n\geq m\\ 0 & \quad\text{otherwise}\end{cases} &
    \tilde{\epsilon}_n&\equiv 0.
    \end{align*}
    \item Replacing $x,y$ by $f(x),f(y)$ (recall that $f$ is bijective) we find that such a constant $C$ exists if and only if there is a constant $C$ with
    $$\left|\frac{g(\{\epsilon_n\})-g(\{\tilde{\epsilon}_n\})}{f(\{\epsilon_n\})-(\{\tilde{\epsilon}_n\})}\right|\leq C$$
    for all $\{\epsilon_n\},\{\tilde{\epsilon}_n\}\in\mathcal{E}$.  But letting
    \begin{align*}
    \epsilon_n&=\begin{cases}1 & \quad\text{if}\ n\geq m\\ 0 & \quad\text{otherwise}\end{cases} &
    \tilde{\epsilon}_n&\equiv 0,
    \end{align*}
    we have
    $$\left|\frac{g(\{\epsilon_n\})-g(\{\tilde{\epsilon}_n\})}{f(\{\epsilon_n\})-(\{\tilde{\epsilon}_n\})}\right|=\frac{3^{m-1}}{2^{m-1}},$$
    which is unbounded as $m\rightarrow \infty$ so that no such constant $C$ independent of $m$ can exist.

    \item Yes.  By parts (a) and (b) we have
    $$\frac{1}{3^{m\eta}}\leq |f(\{\epsilon_n\})-f(\{\tilde{\epsilon}_n\})|^{\eta}$$ for any $\eta>0$ and
    $$|g(\{\epsilon_n\})-g(\{\tilde{\epsilon}_n\})|\leq \frac{1}{2^{m-1}},$$
    so that we have
    $$\frac{|g(\{\epsilon_n\})-g(\{\tilde{\epsilon}_n\})|}{|f(\{\epsilon_n\})-(\{\tilde{\epsilon}_n\})|^{\eta}}\leq 3^{\eta}\left(\frac{3^{\eta}}{2}\right)^{m-1}.$$
    thus, for any $0<\eta <\log(2)/\log(3)$ we have $3^{\eta}/2 < 1$ so that the right side above is bounded for all $m>0$, as required.


\end{enumerate}

\end{enumerate}
\end{document}