1
$\begingroup$

Define

$$\mathbf{f}(s,t) = \begin{pmatrix} (b+a\cos s)\cos t \\ (b+a\cos s)\sin t \\ a\sin s \end{pmatrix},\ 0 < a < b,\ a,b \in \mathbb{R}$$

as the mapping from $\mathbb{R}^2$ to $\mathbb{R}^3$, and let $K$ be the range of $\mathbf{f}$.

Let $g(t) = \mathbf{f}(t,\lambda t)$ be a mapping from $\mathbb{R}^1$ to $K$.

Take $\lambda$ irrational. Show that $g(t)$ is a one-to-one mapping onto a dense subset of $K$.


This problem comes from Baby Rudin (specifically, Chapter 9, problem 12). This is not a homework problem. (It is related to an exam problem from an exam I took almost a year ago).


For the exam, we only needed to show that $g$ was one-to-one. We did not need to show that it was dense in $K$. Showing one-to-oneness is pretty easy. Assume that it is not, then you get a condition where $\cos (t+\lambda 2\pi k) = \cos t$, which cannot hold for any irrational value of $\lambda$.

Is there a clever way to show density, or does it have to just be done directly from the definition?

$\endgroup$
  • 2
    $\begingroup$ Density follows almost immediately from Kronecker's approximation theorem. The theorem can be proved using elementary methods. The general idea of the proof is similar to this answer. Would you consider such an approach satisfactory? $\endgroup$ – Ayman Hourieh Jul 22 '13 at 20:53
  • $\begingroup$ @AymanHourieh Yes, indeed. I did not know that theorem, which is precisely what I was looking for. Perhaps later I should attempt the formal application to this problem. $\endgroup$ – Emily Jul 22 '13 at 21:06
  • $\begingroup$ I've added a formal proof. $\endgroup$ – Ayman Hourieh Jul 22 '13 at 22:14
4
$\begingroup$

Let $f(s_0, t_0)$ be a point on $K$. Consider $g(s_0 + 2\pi n)$ for $n \in \mathbb N$. By Kronecker's approximation theorem, and since $\lambda$ is irrational, we can choose an integer $m$ and a value for $n$ so that:

$$ \left|\frac{t_0 - \lambda s_0}{4\pi} - n \frac{\lambda}{2} + m\right| < \epsilon $$

We have:

\begin{align} \left|\sin t_0 - \sin \lambda(s_0 + 2\pi n)\right| &\le 2 \left|\sin\frac{t_0 - \lambda s_0 - 2\pi n \lambda}{2}\right| \\ &= 2 \left|\sin2\pi\left(\frac{t_0 - \lambda s_0}{4\pi} - n \frac{\lambda}{2} + m\right)\right| \\ &\le 4\pi\left|\frac{t_0 - \lambda s_0}{4\pi} - n \frac{\lambda}{2} + m\right| < 4\pi\epsilon \end{align}

Similarly, we can show that for the same value of $n$: $$ \left|\cos t_0 - \cos \lambda(s_0 + 2\pi n)\right| < 4\pi\epsilon $$

It follows that $g(s_0 + 2\pi n)$ can get arbitrarily close to $f(s_0, t_0)$ as desired. Thus, the image of $g$ is a dense subset of $K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.