Given:
$\alpha\;$ is a $\;\underline{\text{fixed}}\;$ real number.
$\displaystyle \frac{\alpha}{2\pi}\;$ is irrational.
$K(0,1)\;$ denotes the circle of radius 1 centered at the origin.
For $\;z \,\in \,\mathbb{C}
\;\text{and} \;\epsilon \,\in \,\mathbb{R^+}, \;\Delta(z, \epsilon)\;$
denotes
the open disk centered at $\;z\;$ of radius $\;\epsilon.$
$\langle a_n\rangle\;$ denotes the infinite sequence of complex numbers
$\;\{ \,a_1, \,a_2, \,a_3, \cdots \,\}\;$
given by $\;a_n = e^{in\alpha}.$
$z \,\in \,\mathbb{C}\;$ is an accumulation point of %
$\;\langle a_n\rangle \;\Leftrightarrow\;$
$\forall \;\epsilon > 0, \;\Delta(z, \epsilon)\;$ contains $\;a_n\;$ for
infinitely many positive integers $\;n$.
To Prove:
Every element in $\;K(0,1)\;$ is an accumulation point of
$\;\langle a_n\rangle.$
My Request:
The next section gives my attempt to prove this. Is my proof valid?
If not, where did I go wrong?
My Attempt:
Previous flaws corrected. Is the proof now valid?
See also Mohammed M. Zerrak's answer, a much shorter proof.
See also my comment to his shorter proof.
I'm going to explore abandoning the $\text{Arc}(\theta_1, \theta_2)\;$
approach used in this attempt in favor of directly using chord length.
Perhaps that will result in an (alternative) shorter proof.
For any real number $\;\theta,\;$
let $\;P(\theta)\;$ denote $\;\lfloor (\theta/2\pi) \rfloor \;$
(i.e. the floor function).
That is, $\;P(\theta)$ is the largest integer less than
or equal to $\;(\theta/2\pi)\;$.
Let $\;Q(\theta)\;$ denote $\theta \;-\; [2\pi\times P(\theta)]
\;\Rightarrow\; Q(\theta) \,\in \,[0,2\pi).$
Note: Since $\;(\alpha/2\pi)\;$ is irrational,
$\;[Q(\alpha)/(2\pi)]\;$ is also irrational,
which implies that
$\;[Q(\alpha)/(2\pi)] \;\times $ (any integer) is also irrational.
Define an arc function: $\;\text{Arc}(\theta_1, \theta_2)\;$ as follows:
If $\;| \,Q(\theta_1) - Q(\theta_2) \,| \;<\; \pi,\;$ then
$\;\text{Arc}(\theta_1, \theta_2)
\;=\; | \,Q(\theta_1) - Q(\theta_2) \,|.$
Else, $\;\text{Arc}(\theta_1, \theta_2) \;=\;
2\pi \;-\; | \,Q(\theta_1) - Q(\theta_2) \,|.$
As defined, $\;\text{Arc}(\theta_1, \theta_2)\;$ measures the
shortest arc on the unit circle
between the angles $\;\theta_1 \;\text{and} \;\theta_2.$
For $\;k,l \,\in \,\mathbb{Z}, \;\;\theta_1,\theta_2 \,\in \,\mathbb{R},
\;\;Q(\theta_1 + 2k\pi) = Q(\theta_1) \;\;\text{and}
\;\;Q(\theta_2 + 2l\pi) = Q(\theta_2).$
$[E_1]:\;$ Therefore, $\;\text{Arc}(\theta_1 + 2k\pi, \theta_2 + 2l\pi)
\;=\; \text{Arc}(\theta_1, \theta_2).$
$[E_2]:\;$ Therefore, $\;\text{Arc}[Q(\theta_1), Q(\theta_2)]
\;=\; \text{Arc}(\theta_1, \theta_2).$
Given $\;\theta_1,\theta_2,\theta_3 \,\in \,\mathbb{R} <br>
\;\;Q(\theta_1 + \theta_3)\;$ equals either
$\;[Q(\theta_1) + Q(\theta_3)]\;$ or
$\;[Q(\theta_1) + Q(\theta_3) - 2\pi].$
Similar analysis applies when adding $\;\theta_2\;$ to $\;\theta_3.$
Therefore, by combined and repeated use of $\;[E_1]\;$ and $\;[E_2],\;$
$[E_3]:\; \text{Arc}(\theta_1 + \theta_3, \theta_2 + \theta_3)
\;=\; \text{Arc}(\theta_1, \theta_2).$
Also, $\;\forall \;N \,\in \,\mathbb{Z^+}$
$\{ \,Q(N\theta) = N\theta - (2\pi)P(N\theta) \;\;\wedge\;\;
NQ(\theta) = N\theta - N(2\pi)P(\theta) \,\} \;\Rightarrow $
$N[Q(\theta)] \;\equiv\; Q(N\theta) \;(\text{mod} \,2\pi).$
$[E_4]:\;$ Therefore $\;N[Q(\theta)] \;<\; 2\pi \;\Rightarrow\;
N[Q(\theta)] \;=\; Q(N\theta).$
Also, directly from the definition of $\;\text{Arc}(\theta_1, \theta_2),:$
$[E_5]:\: Q(\theta) < \pi \;\Rightarrow\;
\text{Arc}[0, Q(\theta)] = Q(\theta).$
Given a sector of angle $\;\theta$ in the unit circle,
its chord length is $\;2\sin(\theta/2),$ and its arc length is $\;\theta$.
Let $\;f(\theta) \;\equiv\; 2\sin[\theta/2] - \theta \;\Rightarrow\;
f^{\prime}(\theta) \;=\; \cos[\theta/2] - 1.$
Thus, $\;f(0) = 0\;$ and $\;f^{\prime}(0) = 0.$
Also $\;\forall \;\theta \,\in \,(0, \pi/2], \;f^{\prime}(\theta) < 0.$
Therefore, $\;\forall \;\theta \,\in \,[0, \pi/2],\;$ its chord length
is $\;\leq\;$ its arc length.
Therefore, for any $\;\epsilon \,\in \,(0, \pi/2),\;$
if $\;\text{Arc}(\theta_1, \theta_2) \;<\; \epsilon\;$ then the shortest
chord between $\;e^{i\theta_1}\;$ and $\;e^{i\theta_2}\;$
is less than $\;\epsilon,\;$
and consequently, $\;e^{i\theta_2} \,\in \,\Delta(e^{i\theta_1}, \epsilon).$
For a given $\;e^{i\theta} \,\in \,K(0,1),\;$
and $\;\epsilon_0, \epsilon_1 \,\in \mathbb{R^+} \;\ni
\;\epsilon_0 < \epsilon_1,$
$\;\Delta(e^{i\alpha}, \epsilon_0)
\subseteq \Delta(e^{i\theta}, \epsilon_1).$
Therefore, if $\Delta(e^{i\theta}, \epsilon_0)\;$ contains $\;a_n\;$ for
infinitely many
positive integers n, then so will $\Delta(e^{i\theta}, \epsilon_1).$
Consequently, when considering whether a specific value $\;e^{i\theta}\;$ is an accumulation point, the values of $\;\epsilon\;$ considered need not exceed an arbitrary positive number. In this proof, only values of $\;\epsilon\;$ in the range $\;(0, \pi/2)\;$ will be considered. Therefore, by the previous analysis, the function $\;\text{Arc}(\theta_1, \theta_2)\;$ may be used instead of measuring the shortest chord between $\;e^{i\theta_1}\;$ and $\;e^{i\theta_2}.$
That is, for a given $\;e^{i\theta} \,\in \,K(0,1),\;$ and a given
$\;\epsilon \,\in \,(0, \pi/2),$
if there exist an infinite number of positive integers $\;n\;$
such that $\;\text{Arc}(\theta, n\alpha) \;<\; \epsilon,\;$
then there exist an infinite number of positive integers $\;n\;$
such that the shortest chord between
$\;e^{i\theta}\;$ and $\;e^{in\alpha} \;<\; \epsilon,\;$
which implies that there exist an infinite number of positive
integers $\;n\;$
such that $\;a_n \,\in \,\Delta(e^{i\theta}, \epsilon).$
Thus, to show that any element $\;e^{i\theta} \,\in \,K(0,1)\;$
is an accumulation point,
it is sufficient to exhibit an algorithm that may be applied against
any $\underline{\text{fixed}} \;\;\epsilon \,\in (0, \pi/2)\;$
such that:
(1) the algorithm is specific to the specific element
$\;e^{i\theta}\;$ and specific value $\;\epsilon.$
(2) the algorithm generates an infinite strictly increasing sequence of positive
integers $\;\{ \,u_1, \,u_2, \,u_3, \cdots \,\}.\;$
(3) for each positive integer $\;n\;$ in this sequence,
$\;\text{Arc}(\theta, n\alpha) \;<\; \epsilon.$
$\underline{\text{algorithm}}$
Let $\;M\;$ be any positive integer $\;>\; (2\pi/\epsilon)
\;\Rightarrow\; \epsilon \;>\; (2\pi)/M.$
Divide $\;[0,2\pi)$ into $\;M\;$ regions,
each of width $\;(2\pi)/M \;<\; \epsilon.$
Let $\;S_1 \;\equiv\;
\{ \,1, \,2, \,\cdots, \,(M+1) \,\}.$
Consider the values
$\;Q(1\alpha), \,Q(2\alpha), \,\cdots Q([M+1]\alpha).$
Two of these values must fall into the same region.
Therefore, there must be two positive integers
$\;j,k \,\in \,S_1\;: j < k\;$
such that $\;0 \leq\; \left| \,Q(k\alpha) \;-\; Q(j\alpha) \,\right|
\;<\; \epsilon \;<\; (\pi/2) $
This implies that
$0 \leq\; \left| \,[k\alpha - 2\pi P(k\alpha)] \;-\;
[j\alpha - 2\pi P(j\alpha)] \,\right| \;<\; \epsilon \;<\; (\pi/2)
\;\Rightarrow $
$0 \leq\; \left| \,(k - j)\alpha - 2\pi [P(k\alpha) - P(j\alpha)]
\,\right| \;<\; \epsilon \;<\; (\pi/2)
\;\Rightarrow $
$\exists \;\delta \,\in \,[0, \epsilon) \;\ni
(k - j)\alpha - 2\pi [P(k\alpha) - P(j\alpha)]
\;=\; \pm\delta \;\Rightarrow $
$(k-j)\alpha \;\equiv\; \pm\delta \;(\text{mod} \,2\pi) \;\Rightarrow $
$Q[(k-j)\alpha] \;=\; \delta \;\;$ or
$\;\;Q[(k-j)\alpha] \;=\; (2\pi - \delta).$
With $\;k > j,\;$ suppose that
$\;[(k -j)\alpha] \;\equiv\; 0 \;(\text{mod} \,2\pi).$
Then, $\;\exists \;R \,\in \mathbb{Z}\;
\ni [(k -j)\alpha] \;=\; 2\pi R \;\Rightarrow\;
\alpha/(2\pi) \;=\; R/(k-j).$
This is a contradiction, since $\;\alpha/(2\pi)\;$ is assumed irrational.
Therefore, $\;0 \;<\; \delta \;<\; \epsilon \;<\; \pi/2.$
$\underline{\text{Case 1:}\;Q[(k-j)\alpha] \;=\; \delta}.$
If $\;Q(\theta) \;<\; \delta\;$ then set $\;u_1 \;=\;(k-j) \;\Rightarrow\;
\delta \;=\; Q(u_1\alpha) \;\Rightarrow $
$0 \;\leq\; Q(\theta) \;<\; Q(u_1\alpha) \;<\; \epsilon.$
Using $\;E_4, \,E_5,\;$ and $\;E_2,\;$ this implies that
$\text{Arc}[Q(\theta), Q(u_1\alpha)] \;=\;
\text{Arc}[0, Q(u_1\alpha) - Q(\theta)] \;=\; Q(u_1\alpha) - Q(\theta)
\;\Rightarrow $
$\text{Arc}[Q(\theta), Q(u_1\alpha)] \;<\; \epsilon \;\Rightarrow\;
\text{Arc}(\theta, u_1\alpha) \;<\; \epsilon.$
If $\;Q(\theta) \;\geq\; \delta\;$ then set $\;N\;$ equal to the largest
positive integer such that $\;N\delta \leq Q(\theta)\;$
and set $\;u_1 \;=\; N(k-j).$
This will imply that
$\;N\delta \leq Q(\theta) \;<\; (N + 1)\delta.$
Therefore, $\;NQ[(k-j)\alpha] \;\leq\; Q(\theta) \;<\; (N+1)Q[(k-j)\alpha].$
Using $\;E_1\leftrightarrow E_5,\;$ this implies that
$Q(u_1\alpha) \;\leq\; Q(\theta) \;<\; Q(u_1\alpha)
+ \delta \;\Rightarrow $
$\text{Arc}[Q(\theta), Q(u_1\alpha)] \;<\; \epsilon \;\Rightarrow\;
\text{Arc}(\theta, u_1\alpha) < \epsilon.$
$\underline{\text{Case 2:}\;Q[(k-j)\alpha] \;=\; 2\pi - \delta}.$
If $\;2\pi - Q(\theta) \;>\; 2\pi - \delta\;$ then set
$\;u_1 \;=\;(k-j).$
Otherwise, set $\;N\;$ equal to the largest positive integer
such that $(2\pi - N\delta) \geq Q(\theta)\;$
and set $\;u_1\;$ to $\;N(k-j).$
The analysis in Case 2 will then be very similar to Case 1,
so that (again)
$\;\text{Arc}(\theta, u_1\alpha) < \epsilon.$
To compute $\;u_2,\;$ follow the exact same procedure used to compute
$\;u_1\;$ except
let $\;S_2 \;\equiv\;
\{ \,(u_1 + 1), \,(2u_1 + 2), (3u_1 + 3),
\,\cdots, \,(M+1)u_1 + (M+1) \,\}.$
This will guarantee that when the $\;k\;$ and $\;j\;$ are found
that correspond to $\;S_2, \;|k - j| > u_1.$
This will guarantee that the computed $\;u_2\;$ will be greater than
$\;u_1.\;$
This process may be repeated indefinitely, where
$\;S_{(k+1)}\;$ is set to
$\{ \,(u_k + 1), \,(2u_k + 2), (3u_k + 3),
\,\cdots, \,(M+1)u_k + (M+1) \,\}.$
Setting $S_{(k+1)}\;$ as above will guarantee that the
infinite sequence $\;\{ \,u_1, \,u_2, \,u_3, \,\cdots \,\},\;$ will be
strictly increasing.
The procedure itself will guarantee that for each
element $\;u_k\;$ in the sequence,
$\;\text{Arc}(\theta, u_k\alpha) < \epsilon.$
Since this procedure may be followed for any
$\epsilon \,\in \,(0, \pi/2), \;e^{i\theta}\;$ is an accumulation point.
Since this procedure holds for any $\theta, $
every element in $\;K(0,1)\;$ is an accumulation point.
