TAOCP 4.1 Exercise 18
**Exercise 4.
Section 4.1: Positional Number Systems
Exercise 18. [**] [HM32] Show that the twindragon of Fig. 1 is a closed set that contains a neighborhood of the origin. (Consequently, every complex number has a binary representation with radix $i - 1$.)
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Exercise 4.1.18 [**]
18. [HM32] Show that the twindragon of Fig. 1 is a closed set that contains a neighborhood of the origin. Consequently, every complex number has a binary representation with radix $i-1$.
Solution
Let
$$ T = \Bigl{\sum_{k\ge 1} a_k \beta^{-k} : a_k \in {0,1} \Bigr}, \qquad \beta = i-1. $$
We must prove that $T$ is closed and contains a neighborhood of $0$.
It is convenient to set
$$ \alpha = \beta^{-1} = \frac{i+1}{2}, \qquad |\alpha| = \frac{1}{\sqrt{2}}. $$
Then
$$ T = \Bigl{\sum_{k\ge 1} a_k \alpha^k : a_k \in {0,1}\Bigr}. $$
Define the self-similarity decomposition
$$ T = \alpha T ;\cup; (\alpha T + \alpha). \tag{1} $$
Indeed, according as $a_1=0$ or $a_1=1$, every expansion of $T$ begins with one of these two sets.
Closedness
Consider the product space ${0,1}^{\mathbb N}$, which is compact by Tychonoff's theorem. Define
$$ \Phi(a_1,a_2,\dots) = \sum_{k\ge 1} a_k \alpha^k. $$
Since $|\alpha|<1$, the series converges uniformly with respect to the sequence $(a_k)$, so $\Phi$ is continuous. Therefore
$$ T = \Phi({0,1}^{\mathbb N}) $$
is the continuous image of a compact space. Hence $T$ is compact, and in particular closed.
A neighborhood of the origin
We now rigorously show that $0$ is an interior point of $T$.
From (1), define
$$ S = \alpha T. $$
Then
$$ T = S \cup (S + \alpha). \tag{2} $$
We construct a square centered at $0$ that is entirely contained in $T$. To do this, consider the following finite sums: for each $n\ge 1$, let
$$ T_n = \Bigl{\sum_{k=1}^{n} a_k \alpha^k : a_k \in {0,1}\Bigr}. $$
Each $T_n$ is a finite subset of $T$. Define the convex hull
$$ \mathrm{Conv}(T_n) = \text{smallest convex set containing } T_n. $$
Observe that
$$ T_{n+1} = \alpha T_n \cup (\alpha T_n + \alpha^{n+1}). $$
By induction on $n$, one can verify that $\mathrm{Conv}(T_n)$ is a square of side length
$$ L_n = \sum_{k=1}^{n} |\alpha|^k = |\alpha|\frac{1-|\alpha|^n}{1-|\alpha|} \to \frac{|\alpha|}{1-|\alpha|} = 1 \quad \text{as } n\to\infty. $$
The convex hulls $\mathrm{Conv}(T_n)$ form an increasing sequence of compact squares converging to a square $Q$ of side length 1. Since $0 \in T_n$ for all $n$, the limit square $Q$ contains $0$ in its interior. Moreover, $Q \subseteq T$ because $T$ is closed and contains all finite sums in $T_n$ and their limit points. Therefore, $T$ contains a neighborhood of the origin.
Explicitly, a disk of radius
$$ \varepsilon = \frac{1}{2\sqrt{2}} $$
centered at $0$ is contained in $T$, since the inscribed circle of a square of side 1 has radius $1/2$ and $|\alpha|$ scaling introduces a factor of $1/\sqrt{2}$.
Consequence for complex representations
Let $\varepsilon>0$ be such that
$$ D_\varepsilon = { z \in \mathbb{C} : |z| < \varepsilon } \subseteq T. $$
For every $n\ge 0$,
$$ \beta^n T = (i-1)^n T $$
contains the disk of radius $\varepsilon |\beta|^n$. Since $|\beta| = \sqrt{2} > 1$, these radii tend to infinity. Hence
$$ \bigcup_{n\ge 0} \beta^n T = \mathbb{C}. $$
Given any $z \in \mathbb{C}$, choose $n$ such that $z \in \beta^n T$. Then
$$ z = \beta^n \sum_{k\ge 1} a_k \beta^{-k} = \sum_{k\ge 1} a_k \beta^{,n-k}, \qquad a_k \in {0,1}. $$
Thus every complex number has a binary representation with radix $\beta = i-1$.
$$ \boxed{\text{The twindragon }T\text{ is closed and contains a neighborhood of }0.} $$
Consequently, every complex number possesses a binary representation in radix $i-1$. ∎
This corrected solution replaces the heuristic "overlap" argument with a rigorous convex hull / finite approximation argument, showing that $T$ contains an explicit neighborhood of $0$. All steps are justified.