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

Baby Rudin ch9 q12d

Fix two real numbers $$a$$ and $$b$$, $$0 . Define a mapping $$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.

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 $$0. The number $$(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.
• sounds similar to $\{m\lambda \}$ is dense in $[0, 1)$. I should have been able to see the similarity since both involve density and an irrational $\lambda$. Thanks! – Vinay Deshpande Oct 27 '20 at 14:53