2

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?

Emily
  • 35,688
  • 6
  • 93
  • 141

1 Answers1

5

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 $K$.

Ayman Hourieh
  • 39,603