Find out whether $M:=\{\left(x\cdot\cos\left(\frac{1}{x}\right),\,x\cdot\sin\left(\frac{1}{x}\right)\right)|\,x>0\}$ is a submanifold of $\mathbb{R}^2$ or not.
My guess is that it's not since I've tried to construct a homeomorphism from $\mathbb{R}$ to $M$ $\left(x\mapsto \left(x\cdot\cos\left(\frac{1}{x}\right),\,x\cdot\sin\left(\frac{1}{x}\right)\right)\right)$ and failed in proving bijectivity. I then tried to construct multiple charts from $\mathbb{R}$ and $\mathbb{R}^2$ but it hasn't worked out so far because of the way $M$ looks near the origin.
Thank you very much in advance.
It is a submanifold.
Consider the map $f : (0,\infty) \to \mathbb R^2, f(x) = (x\cos(\frac{1}{x}),x\sin(\frac{1}{x}))$. We have $M = f((0,\infty))$. The map $f$ is injective since $\lVert f(x) \rVert = \sqrt{x^2\cos^2(\frac{1}{x}) + x^2\sin^2(\frac{1}{x})} = \sqrt{x^2} = x$ which implies $f(x_1) \ne f(x_2)$ for $x_1 \ne x_2$. Thus $\tilde f : (0,\infty) \stackrel{f}{\to} M$ is a bijection. Moreover, $g = \lVert - \rVert_M : M \to (0,\infty)$ is continuous and we have $g \circ \tilde f = id$, therefore $\tilde f$ is homeomorphism.
We have $f'(x) = (\cos(\frac{1}{x}) + \frac{1}{x}\sin(\frac{1}{x}), \sin(\frac{1}{x}) - \frac{1}{x}\cos(\frac{1}{x}))$ and we get $$\lVert f'(x) \rVert = \sqrt{\cos^2(\frac{1}{x}) + \frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x}) + \frac{1}{x^2}\sin^2(\frac{1}{x}) + \sin^2(\frac{1}{x}) - \frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x}) + \frac{1}{x^2}\cos^2(\frac{1}{x})} \\ = \sqrt{1 + \frac{1}{x^2}} \ne 0 .$$ Thus $f'(x) \ne 0$ so that $f$ is an immersion. Hence $f$ is a smooth embedding and $M$ is a smooth submanifold of $\mathbb R^2$.