I don't think this is meant to be hard. I came up with this set in $\mathbb{R}^3$ in order to address this bigger problem I am doing. I took the helicoid and thickened it in the direction of a transversal vector at each point: If $f(r, \theta) = (r \cos \theta, r \sin \theta, \theta)$ where $(r, \theta) \in \mathbb{R}^2$ then I define this set $U := \{f(r, \theta) + q(\sin \theta, -\cos \theta, r) | q \in (-\epsilon, \epsilon)\}$. I want to show $U$ is an open set in $\mathbb{R}^3$! I'm having a stupid moment, could someone please help?
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I mistook helicoid for a helix in a now deleted comment. Sorry about not checking the meaning. – Jyrki Lahtonen Apr 01 '18 at 09:26
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You have $$f_r=(\cos\theta,\sin\theta,0),\qquad f_\theta=(-r\sin\theta,r\cos\theta,1)\ ,\qquad f_r\times f_\theta=(\sin\theta,-\cos\theta, r)$$ (note the $r$ at the end of the line!). The unit normal is therefore given by $$n(r,\theta)={1\over\sqrt{1+r^2}}(\sin\theta,-\cos\theta, r)\ .$$ A helicoidal shell of thickness $2\epsilon$ is then produced by $$\Psi:\quad(r,\theta,t)\mapsto f(r,\theta)+ t\,n(r,\theta)\qquad\bigl(r\in{\mathbb R},\ \theta\in{\mathbb R},\ -\epsilon<t<\epsilon\bigr)\ .$$ In order to show that this shell is an open set in ${\mathbb R}^3$ you have to verify that the Jacobian $J_\Psi(r,\theta, t)$ is $\ne0$ at all parameter points $(r,\theta,t)$ with $|t|$ sufficiently small.
Christian Blatter
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Why would the Jacobian being nonzero mean that the shell is open, sorry? – Acton Apr 01 '18 at 09:22
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@ActonGrey inverse function theorem https://math.stackexchange.com/questions/263837/smooth-map-with-surjective-jacobian-is-open – Calvin Khor Apr 01 '18 at 10:05
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@Christian Blatter typos: the third component of $f_\theta$ should read $1$, not $r$; and "open set in $\mathbb{R}$" should read "open set in $\mathbb{R}^3$" – xFioraMstr18 May 15 '18 at 11:37
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and @Acton uses a shell produced by $t\sqrt{1+r^2}n(r,\theta)$, not $tn(r,\theta)$ – xFioraMstr18 May 15 '18 at 11:54
