Showing that the function $C(b)$ is a compact set for $|b| > 1$ I am reading "An Invitation to Dynamical Systems", and one of the challenge problems is to prove that $C(b)$ is a compact set where $C(b)$ is defined as the set of all numbers that can be expressed in the form $0.d_1d_2d_3..._b$ (base $b$) where the $d$'s are $0$'s or $1$'s, and where $|b| > 1$. The book defines a compact set as a set which is both bounded and closed, so I suppose I need to prove both these properties, but I don't even know where to begin. Any help would be appreciated.
Edit: Here is what I mean when I say that $b$ can be any number, real of complex. This is the way the author defines a base: With $x = 0.d_1d_2d_3..._b$, we can say that $$x = \sum_{k = 0}^{\infty} \frac{d_k}{b^k}.$$
This definition works for complex $b$ as well.
 A: It seems the following. 
If $|b|\le 1$ then in order to converge, the series for $x$ should have only finitely many non-zero coefficients $d_k$. Moreover, for each $b$ the set $\Bbb N_b=\{k\in\Bbb N:\operatorname{arg} b^{-k}\in [-\pi/3;\pi/3]\}$ is infinite. Put $x_n=\sum_{k\in\Bbb N_b\cap [1,n]} \frac{1}{b^k}$. Since $\operatorname{Re} b^{-k}\ge |b^{-k}|/2=|b|^{-k}/2\ge 1/2$ for each number $k\in \Bbb N_b$, we see that the sequence $\{x_n\}\subset C(b)$ is unbounded, so the set $C(b)$ is not compact.
So we suppose that $|b|>1$.
To prove that the set $C(b)$ is compact it suffices to show that the set $C(b)$ is a continuous image $f(\mathcal C)$ of the Cantor cube $\mathcal C=\{0,1\}^\Bbb N$ endowed with the topology of Tychonoff  product. The way to determine the map $f$ is natural. For each point  $d=\{d_k\}\in\mathcal C$ we put 
$$f(d)= \sum_{k = 0}^{\infty} \frac{d_k}{b^k}.$$
Cantor cube $\mathcal C$ has a standard metric $\rho$ such that $\rho(d,d’)=2^{-l}$, where $d,d’\in\mathcal C$, $d=\{d_k\}$, $d’=\{d’_k\}$, and $l$ is a maximal number such that $d_k=d_k’$ for each $k\le l$. Let $d=\{d_k\}$, $d’=\{d’_k\}$ be arbitrary points of the space $\mathcal C$  such that $\rho(d,d’)\le 2^{-l}$. Then $d_k=d_k’$ for each $k\le l$. Hence  
$$\left|f(d)-f(d’)\right|=\left| \sum_{k = 0}^{\infty} \frac{d_k}{b^k}- \sum_{k = 0}^{\infty} \frac{d’_k}{b^k} \right|=\left| \sum_{k = 0}^{\infty} \frac{d_k-d’_k}{b^k}\right|=\left| \sum_{k=l+1}^{\infty} \frac{d_k-d’_k}{b^k}\right|\le  \sum_{k=l+1}^{\infty} \frac{|d_k-d’_k|}{|b^k|}\le  \sum_{k=l+1}^{\infty} \frac{2}{|b|^k}=2\frac{|b|^{-l-1}}{1-|b|^{-1}}.$$
Since $\lim_{l\to\infty}\frac{|b|^{-l-1}}{1-|b|^{-1}}=0$, we see that the function $f$ is (uniformly) continuous.
PS. For me it is a usual way to prove compactness (see, for instance, here :-) ). Compacts which are images of Cantor  cubes are called dyadic. They comprise a big enough class. For instance, each non-empty metrizable compact and each compact (Hausdorff) topological group is dyadic. 
A: Since we can think of $x$ as a literal number, we have 
$|x|=\left|\sum_0^\infty \frac{d_k}{b^k} \right| \le \sum_0^\infty \frac{d_k}{|b^k|} \le \sum_0^\infty \frac{1}{|b^k|} = \frac{1}{1-|b|}$,
since I believe you meant that $|b| > 1$. Here I am using the literal notion of metric of numbers $d(x,y)=|x-y|$. Since $d_k \le 1$, I get the above upper bound on the norm of $x$, so this gives you boundedness of $C(b)$. I'm not sure off the top of my head how to get closedness by a similar "direct" method (but it is probably possible).
I should mention that the perspective discussed in the comments that $C(b)$ is the continuous image of $\{0,1\}^\mathbb{N}$ is very slick and worth thinking about. Of course, you'll probably want to use Tychanoff's theorem to show that $\{0,1\}^\mathbb{N}$ is compact.
A: Closed
Observe that $C(b)$ is an intersection of closed sets:
$$
\begin{align*}
C(b) &= \bigcap_{k=1}^{\infty}C_k(b),\\
\end{align*}
$$
where $C_k(b)$ is the set where $d_1,\cdots,d_k\in \{0,1\}$. $C_k(b)$ is closed as it is a finite union of intervals.
Bounded
$$|x|\leq \sum_{k=1}^{\infty}\frac{1}{|b|^k}=\frac{1}{|b|-1}<\infty$$
