Kronecker Approximation

After class one day, one of my students shared with me about this fascinating result. The source material comes from this video, and the writeup is my interpretation of its key ideas.

Let $\mathbb K$ be an ordered field.

Problem 1. Prove that for $\mathbb K$ -sequences $( x_n ), ( y_n )$ , if $x_n \to 0$ and $y_n$ is bounded, then $x_n \cdot y_n \to 0$ .

(Click for Solution)

Solution. Since $(y_n)$ is bounded, find $M > 0$ such that $|y_n| < M$ . Fix $\epsilon > 0$ . Since $x_n \to 0$ , for any $k > 0$ , there exists $N \in \mathbb N$ such that if $n > N$ , then $|x_n| < k \cdot \epsilon$ . By construction,

$|x_n \cdot y_n| = |x_n| \cdot | y_n | < k \cdot \epsilon \cdot M.$

Setting $k := 1/M > 0$ yields the result $x_n \cdot y_n \to 0$ .

For the rest of this exercise, let $\alpha$ be an irrational number.

Definition 1. For any $x \in \mathbb R$ , define the fractional part of $x$ by $\{ x \} := x - \lfloor x \rfloor$ .

Problem 2. Prove that for distinct positive integers $p, q$ , $\{p \alpha\} \neq \{q \alpha\}$ .

(Click for Solution)

Solution. Suppose for a contradiction that there exist distinct positive integers $p, q$ such that $\{p \alpha\} = \{q \alpha \}$ . By the definition of the fractional part,

$\begin{aligned}p \alpha - \lfloor p\alpha \rfloor &= q \alpha - \lfloor q \alpha \rfloor\quad \Rightarrow \quad \alpha = \frac{ \lfloor p\alpha \rfloor - \lfloor q \alpha \rfloor}{p - q} \in \mathbb Q, \end{aligned}$

a contradiction.

Problem 3. Prove that for any $n \in \mathbb N$ , there exist integers $p_n, q_n$ such that

$0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

(Click for Solution)

Solution. Fix $n \in \mathbb N$ . By Problem 2, consider the $n+1$ distinct values

$\{\alpha \},\{2\alpha\},\dots, \{n \alpha\}, \{(n+1)\alpha\}$

and the $n$ equally-spaced sub-intervals of $[0, 1)$ :

$[0, 1/n), [1/n, 2/n), \dots, [(n-1)/n, 1).$

This means there exists integers $i_n, p_n, q_n$ such that

$\{p_n \alpha\}, \{q_n\alpha\} \in [i_n/n, (i_n+1)/n)\quad \Rightarrow \quad 0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

Problem 4. Define

$A := \mathbb Z + \alpha \mathbb Z \equiv \{ m + n \alpha : m, n \in \mathbb Z\}.$

Prove that for any $x \in \mathbb R$ , there exists an $A$ -sequence $( x_n )$ such that $x_n \to x$ .

(Click for Solution)

Solution. Fix $x \in \mathbb R$ . Use Problem 3 to define the sequence $(y_n)$ by $y_n := \{ p_n \alpha \} - \{ q_n \alpha \}$ . It is clear that $y_n \to 0$ and each $y_n \neq 0$ . Define $x_n := y_n \cdot \lfloor x/y_n \rfloor$ . Then

$\displaystyle x_n = y_n \cdot \lfloor x/y_n \rfloor = y_n \cdot \left( \frac x{y_n} - \left\{ \frac x{y_n} \right\}\right)= x - y_n \left\{ \frac x{y_n} \right\}.$

By the definition of the fractional part, the sequence $( \{x / y_n \} )$ is bounded. Since $y_n \to 0$ , by Problem 1, we have $y_n \cdot \{x / y_n \} \to 0$ . Hence, $x_n \to x$ .

It remains to verify that $x_n \in A$ for each $n$ . Indeed,

$\begin{aligned} x_n &= y_n \cdot \lfloor x/y_n \rfloor \\ &= (\{ p_n \alpha \} - \{ q_n \alpha \}) \cdot \lfloor x/y_n \rfloor \\ &= \{ p_n \alpha \} \cdot \lfloor x/y_n \rfloor - \{ q_n \alpha \} \cdot \lfloor x/y_n \rfloor \\ &= (p_n \alpha - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor - (q_n \alpha - \lfloor q_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor \\ &= \underbrace{ (\lfloor q_n \alpha \rfloor - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor }_{\in \mathbb Z} +\, \alpha \cdot \underbrace{ (p_n - q_n) \cdot\lfloor x/y_n \rfloor }_{\in \mathbb Z} \\ &\in \mathbb Z + \alpha \mathbb Z = A, \end{aligned}$

as required.

Problem 5. Define

$B := \sin(\mathbb N) \equiv \{ \sin(n) : n \in \mathbb N\}.$

Prove that for any $y \in [-1, 1]$ , there exists a $B$ -sequence $( y_n )$ such that $y_n \to y$ .

(Click for Solution)

Solution. For any $y \in [-1, 1]$ , find $u \in (0, \infty) \subseteq \mathbb R$ such that $\sin(u) = y$ . Define

$C := \mathbb Z + 2\pi \mathbb Z.$

By Problem 4, there exists a $C$ -sequence $(u_n)$ such that $u_n \to u$ . Denote $u_n = s_n + 2\pi \cdot t_n$ with $s_n \in \mathbb N$ . Defining $y_n = \sin(s_n) \in B$ ,