Every point of the unit circle is a limit point of $x_k = \left(1+\frac{1}{2^k}\right)\begin{bmatrix} \cos k\\\sin k\\ \end{bmatrix}$

Question

Consider the function $f:\mathbb{R}^n\to \mathbb{R}$ defined by $f(x)
= |x|^2$

Show that $f(x_{k+1})<f(x_{k})$ for

$$x_k = \left(1+\frac{1}{2^k}\right)\begin{bmatrix}
\cos k \\
\sin k\\
\end{bmatrix}$$

Show also that every point on the unit circle is a limit point of
$\{x_k\}$. Hint:

Every value $\theta\in[0,2\pi]$ is a limit point of the subsequence
$\eta_k$ defined by $$\eta_k = k \mod \ 2\pi$$

For the first question I simply did $$x_k = \begin{bmatrix}
\left(1+\frac{1}{2^k}\right)\cos k \\
\left(1+\frac{1}{2^k}\right)\sin k\\
\end{bmatrix} \implies |x_k|^2 = \left(1+\frac{1}{2^k}\right)^2\cos^2k + \left(1+\frac{1}{2^k}\right)^2 \sin^2 k = \left(1+\frac{1}{2^k}\right)^2$$

Then it's immediate that

$$\left(1+\frac{1}{2^{k+1}}\right)^2< \left(1+\frac{1}{2^k}\right)^2$$

For the second question, I first noted that it looks like $\{\eta_k\}$ 'spans' the entire interval $[0,2\pi]$. Since every point $p$ in the unit circle is parametrized by $p = (\cos \theta_0, \sin\theta_0)$ for some $\theta_0\in[0,2\pi]$, I can think of this $\theta_0$ as a limit point of $\eta_k$. That is, given $\epsilon>0$, there's always $k\in \mathbb{N}^+$ such that $|\eta_k-\theta_0|<\epsilon$.

I must show that for every point $p = (\cos\theta_0,\sin\theta_0)$ in the circle, given $\epsilon>0$, there's always $k\in\mathbb{N}^+$ such that $|p-x_k|<\epsilon$

I think it has something to do with finding a continuous map between an open of $[0,2\pi]$ and the unit circle, and since at $[0,2\pi]$ the sequence is always close to it, then at the image it will be close to the unit circle. Something like that.

So given a point $p$ in the unit circle, I know there's a continuous map from $[0,2\pi]$ to the unit circle. Therefore, given an open $V$ around my point $p$, I'll have an open $U$ around some $\theta_0$. I know that this open includes $\eta_k$ for some $k$. And I also know that $U\subset F^{-1}(V)$, that is, $F(U)\subset V$. Now I need to reason about $F(U)$. No matter how small $V$ is, $F(U)$ will always be in $V$. That is, the image of both $\theta_0$ and $\eta_k$ will be close. But it's the image under the function of the unit circle. I don't think it gives me anything useful.

Answer 1

Choose a point $(\cos\theta,\sin\theta)$ on the circle. Let $\varepsilon>0$. Applying the hint, there exists a subsequence of integers $n_k$ such that $\theta=\lim_{k\to\infty}(\hbox{$n_k$ mod $2\pi$})$. Then the sequence of points $x_{n_k}$ converges to $(\cos\theta,\sin\theta)$, by the continuity of the trigonometric function and the fact that $(1+2^{-n_k})$ tends to $1$. In fact $$x_{n_k}=(1+2^{-n_k})(\cos n_k,\sin n_k)=(1+2^{-n_k})(\cos(\hbox{$n_k$ mod $2\pi$}),\sin(\hbox{$n_k$ mod $2\pi$})),$$ which tends to $(\cos\theta,\sin\theta)$ as $k\to\infty$.