I was reading a proof of Kronecker's theorem, which is: Let $\theta$ be an irrational number. For all real $\alpha$ and all $\epsilon >0 $, there exist integers $a,c$ with $|a\theta - \alpha - c| < \epsilon$.
The proof uses Dirichlet's theorem, whereby for all $\epsilon > 0$, there exist integers $a,b$ with $|a \theta - b| < \epsilon$. Since $\theta$ is irrational, $0 < |a \theta - b|$. Then the series of points $0,\{a \theta\},\{2a\theta\},...$ form a chain across the interval $[0,1)$, whose mesh is $< \epsilon$. The chain goes left to right if $a \theta - b > 0$, and right to left o.w.. Thus $\{ \alpha \}$ falls between links, and there are integers $a,c$ with:
$$|a\theta - \alpha - c| < \epsilon$$
I don't understand the portion:
"... Then the series of points $0,\{a \theta\},\{2a\theta\},...$ form a chain across the interval $[0,1)$, whose mesh is $< \epsilon$. The chain goes left to right if $a \theta - b > 0$, and right to left o.w...."
How do we get to conclude using Dirichlet that $0,\{a \theta\},\{2a\theta\},...$ form a chain across the interval $[0,1)$, and that the chain goes left to right if $a \theta - b > 0$, and right to left o.w...."
Any help ?
You have, for some integer $c$ that
$$a\theta = c + \{a\theta\} \tag{1}\label{eq1A}$$
where $\{a\theta\}$ is the fractional part of $a\theta$. You then get that
$$\left|a\theta - b\right| \lt \epsilon \implies \left|(c-b) + \{a\theta\}\right| \lt \epsilon \implies -\epsilon \lt (c-b) + \{a\theta\} \lt \epsilon \tag{2}\label{eq2A}$$
Since $0 \lt \{a\theta\} \lt 1$, any $\epsilon \lt 1$ means $c - b = 0$ or $c - b = -1$. The first case gives
$$a\theta - b \gt 0 \; \text{ and } \; 0 \lt \{a\theta\} \lt \epsilon \tag{4}\label{eq4A}$$
Note for any positive integer $k \lt \frac{1}{\epsilon} \implies k\epsilon \lt 1$ that \eqref{eq4A} gives $0 \lt k\{a\theta\} \lt k\epsilon \lt 1$, with \eqref{eq1A} then giving
$$ka\theta = kc + k\{a\theta\} \implies \{ka\theta\} = k\{a\theta\} \tag{5}\label{eq5A}$$
This means going from $\{ka\theta\}$ to $\{(k+1)a\theta\}$ involves a step, i.e., mesh size, of $\{a\theta\} \lt \epsilon$, with it going from left to right, i.e., increasing from $0$ towards $1$.
The second case of $c - b = -1$ gives
$$a\theta - b \lt 0 \; \text{ and } \; -\epsilon \lt -1 + \{a\theta\} \lt 0 \tag{6}\label{eq6A}$$
For any positive integer $k \lt \frac{1}{\epsilon} \implies k\epsilon \lt 1$ you thus get
$$\begin{equation}\begin{aligned} -k\epsilon & \lt k(-1 + \{a\theta\}) \lt 0 \\ -1 & \lt k(-1 + \{a\theta\}) \lt 0 \\ 0 & \lt 1 + k(-1 + \{a\theta\}) \lt 1 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Using \eqref{eq1A} now gives for these $k$ that
$$\begin{equation}\begin{aligned} a\theta & = c + 1 - 1 + \{a\theta\} \\ ka\theta & = kc + k + k(-1 + \{a\theta\}) \\ ka\theta & = kc + k - 1 + (1 + k(-1 + \{a\theta\})) \\ \{ka\theta\} & = 1 + k(-1 + \{a\theta\}) \end{aligned}\end{equation}\tag{8}\label{eq8A}$$
Instead of starting at $0$, if you start at $1$ for $k = 0$ instead (this is something the proof perhaps have made more clear as claiming to always start at $0$ is misleading), then each increase in $k$ decreases the value by $1 - \{a\theta\} \lt \epsilon$, i.e., the step size is less than $\epsilon$, with it going from right to left in this case, i.e., decreasing from $1$ towards $0$.