Exercise 3.19

Let $A$ denote the set of all $\alpha$ sequences as defined in the text and let $x(A)=\{x(\alpha ):\alpha \in A\}$.

For a given sequence $\alpha \in A$ let

We first show that $x(\alpha )=\sum <!--\; FUNCTION\; APPLICATION\; -->{\beta }_n$ converges. So consider any $\alpha \in A$. If $\{{\alpha }_n\}$ ends in all zeros (i.e. there is an $N\ge 1$ where ${\alpha }_n=0$ for all $n\ge N$) then clearly $\sum <!--\; FUNCTION\; APPLICATION\; -->{\beta }_n$ is effectively finite and so therefore converges. If this is not the case then $\{{\alpha }_n\}$ clearly has a subsequence $\{{\alpha }_{k_n}\}$ where ${\alpha }_{k_n}=2$ for all $n\ge 1$. Then, noting that obviously ${\beta }_n\ge 0$, we have

where we have invoked Theorem 3.20b. It follows then from the Root Test (Theorem 3.33) that $\sum <!--\; FUNCTION\; APPLICATION\; -->{\beta }_n$ converges.

Now, for a sequence $\alpha \in A$, $x(\alpha )$ is the real number whose representation in base three is $0.{\alpha }_1{\alpha }_2{\alpha }_3\dots$, i.e. ${\alpha }_n$ is the $n$th digit after the decimal point. Clearly if ${\alpha }_n=0$ for all $n\ge 1$ then $x(\alpha )=0$. We also note that if ${\alpha }_n=2$ for all $n\ge 1$ then we have

analagously to the way in which $0.99999\dots =1$. It follows then that $x(A)\subseteq [0,1]$.

First consider any $m\ge 1$ for any $\alpha \in A$. We then have

where we adopt the convention that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=p}^q=0$ for $p>q$, which is needed for the first sum if $m=1$. Continuing, we have

where we define

noting that for the sum

so that $j_{\alpha ,m}$ is an integer (since the ${\alpha }_n$ are). Note also that $j_{\alpha ,1}=0$, following the convention. Now we examine the last sum in (1). Letting $l=n-m$ so that $n=l+m$ we let

This sum of course depends on $\alpha$ but we can find its bounds. Clearly it will be the lowest when ${\alpha }_{l+m}=0$ for $l\ge 1$ and the largest when ${\alpha }_{l+m}=2$ for $l\ge 1$. We therefore have

Recombining this with (1) we have

where $j_{\alpha ,m}\in \mathbb{N}$ (where $0\in \mathbb{N}$) and $0\le {\delta }_{\alpha ,m}\le 1$.

Moving along, let $P$ denote the Cantor set. Now, by equation (2.24) in the text

where ${\mathbb{Z}}^{\text{+}}$ denotes the set of positive integers. Note that the text says that $k\in {\mathbb{Z}}^{\text{+}}$ but this should really be $k\in \mathbb{N}$ as written above. This is logically equivalent to

Finally we show that $x(A)=P$.

( $\subseteq$) Consider any $\alpha \in A$ and consider any $m\in {\mathbb{Z}}^{\text{+}}$ and $k\in \mathbb{N}$. If ${\alpha }_m=0$ then by (2)

So in the sub-case where $j_{\alpha ,m}>k$ we have $j_{\alpha ,m}\ge k+1$ so that

In the other sub-case when $j_{\alpha ,m}\le k$ we have

On the other hand if ${\alpha }_m=2$ then again by (2)

So in the sub-case where $j_{\alpha ,m}\ge k$ we have

and the other sub-case where $j_{\alpha ,m}<k$ we we have $j_{\alpha ,m}+1\le k$ so that

Thus, since the cases are exhaustive, we have shown that the right side of (3) is true, from which it follows that $x(\alpha )\in P$. Since $\alpha$ was an arbitrary sequence, $x(A)\subseteq P$.

( $\supseteq$) Consider any $y\in P$. Then clearly $y\in [0,1]$. So let $\alpha$ be the sequence corresponding to the base three representation of $y$ as described above; thus $y=x(\alpha )$. If $\alpha$ contains only zeros and twos, i.e. ${\alpha }_n\in \{0,2\}$ for all $n\ge 1$ then clearly $\alpha \in A$ so that $y\in x(A)$. So suppose that there is an $n\ge 1$ where ${\alpha }_n=1$. We will show that this is the only digit that is one and that, furthermore, there is an ${\alpha }^{\prime }$ such that ${\alpha }^{\prime }\in A$ and $x({\alpha }^{\prime })=x(\alpha )=y$.

So let $m$ be the index of the first digit that is a one, i.e. $m=\min <!--\; FUNCTION\; APPLICATION\; -->\{k\in {\mathbb{Z}}^{\text{+}}:{\alpha }_k=1\}$. Then since $y\in P$, $m\in {\mathbb{Z}}^{\text{+}}$, and $j_{\alpha ,m}\in \mathbb{N}$ we have $3^{m}y\le 3j_{\alpha ,m}+1$ or $3^{m}y\ge j_{\alpha ,m}+2$ by (3).

If $3^{m}y\le 3j_{\alpha ,m}+1$ then since $y=x(\alpha )$ we have by (2)

Since also we know that ${\delta }_{\alpha ,m}\ge 0$ it has to be that ${\delta }_{\alpha ,m}=0$, which implies that every digit after ${\alpha }_m$ is zero so that ${\alpha }_m$ is the only digit that is one. But then we can form ${\alpha }^{\prime }$ where

Clearly $x({\alpha }^{\prime })=x(\alpha )=y$ and ${\alpha }^{\prime }\in A$.

If $3^{m}y\ge 3j_{\alpha ,m}+2$ then we again have by (2)

Since we also know that ${\delta }_{\alpha ,m}\le 1$ it has to be that ${\delta }_{\alpha ,m}=1$, which means that ${\alpha }_k=2$ for $k>m$, i.e. the base three representation of $x(\alpha )$ ends in infinitely many twos after the one. Then we can construct a terminating ${\alpha }^{\prime }$ where

Again clearly $x({\alpha }^{\prime })=x(\alpha )=y$ and ${\alpha }^{\prime }\in A$.

Thus in either case there is an ${\alpha }^{\prime }\in A$ where $y=x(\alpha )=x({\alpha }^{\prime })$ so that $y\in x(A)$. Since $y$ was arbitrary this shows that $P\subseteq x(A)$, which completes the proof since it has been shown that $x(A)=P$.