To prove image of a subset of a set is dense in image of the set

Question

Baby Rudin ch9 q12d

Fix two real numbers $a$ and $b$, $0 <a < b$. Define a mapping $f = (f_1,f_2 , f_3)$ of $\mathbb{R}^2$ into $\mathbb{R}^3$ by

$f_1(s, t) = (b + a \cos s) \cos t$

$f_2(s, t) =(b + a \cos s) \sin t$

$f_3(s, t) = a \sin s$.

Describe the range $K$ of $f$. (It is a certain compact subset of $\mathbb{R}^3$.)

(d) Let $\lambda$ be an irrational real number, and define $g(t) = f(t, \lambda t)$. Prove that $g$ is a 1-1 mapping of $\mathbb{R}^1$ onto a dense subset of $K$.

I proved 1-1.

Let $L=\{(t, \lambda t)\mid t\in \mathbb{R}\}$

To prove $img\, g$ is dense in $K$.

i.e. to prove $f (L)$ is dense in $K$

attempt 1: $B\subset A$ is dense in $A$ and $f$ is continuous on $A$ $\implies$ $f(B)$ is dense in $f(A)$.

But $L$ is not dense in $\mathbb{R}^2$ $\implies$ this attempt fails.

attempt 2: To prove $\forall \, \epsilon >0, \, \forall \, (s, t) \in \mathbb{R}^2, \, \exists \, t' \in \mathbb{R}$ s.t. $d(f(s,t), f(t', \lambda t'))< \epsilon$. This is probably tough.

1-1ness is not used yet.

Please give a hint. Please do not give solution. Thanks!

Answer 1

To prove density here, you try to show that for $\theta\in\mathbb{R}$ and any $\varepsilon>0$, $\exists m,n\in\mathbb{Z}$ s.t. $$|2\pi n\lambda-2\pi m-\theta|<\varepsilon$$

An equivalent version of the above and try to prove this: for any $\varepsilon'>0$ and any real number $c\in\mathbb{R}$. $\exists m,n\in\mathbb{Z}$ s.t. $$|n\lambda-m-n|<\varepsilon'$$

Take $r+1$ numbers $\{0,\lambda-[\lambda],\ldots, r\lambda-[r\lambda]\}\subset[0,1)$. Two of them must be less than $\frac{1}{r}$, say $0<s\lambda-[s\lambda]-t\lambda+[t\lambda]<\frac{1}{r}$. The number $(s-t)\lambda$ lies within $\frac{1}{r}$ of an integer, $(s-t)\lambda=k+\delta$ for $0<\delta<\frac{1}{r}$. Let $p$ be the unique integer satisfying $p\delta\leq c<(p+1)\delta$. Can you complete the rest of the proof of this statement?

Try to prove this statement, then you can get the one with $\theta$.

My method does not typically evolve the result that $g(t)$ is one-to-one.