TITLE: Showing that the function $C(b)$ is a compact set for $|b| > 1$
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: 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
$$|x|\leq \sum_{k=1}^{\infty}\frac{1}{|b|^k}=\frac{1}{|b|-1}<\infty$$
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