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let $f,g : \mathbb{R} \rightarrow \mathbb{R}$ be $C^1$ with $f(x),g(x),f'(x),g'(x)>0$ for every $x$. show that $\forall a,b \in \mathbb{R}$ the set $$\{ (x,y) \in \mathbb{R^2}:f(x)\cos y =a , g(x)\sin y=b\}$$ has no cluster points.

My attempt

let $F(x,y) = f(x)\cos y -a$ and $G(x,y) = g(x)\sin y -b$ with \begin{vmatrix} F_x & F_y \\ G_x & G_y \notag \end{vmatrix} $= f'(x)g(x)\cos^2x + g'(x)f(x)\sin^2y >0$

But how does one imply that the set has no cluster points?

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    For each $a$ and $b$ this set is the preimage of a closed set under a continuous function hence closed. Thus, all cluster points belong to the set. You have shown that at any point $(x,y)$ in the set the determinant is non-zero and the function $F$ is $C^{1}$. So by the inverse function theorem there is a neighbourhood of $(x,y)$ in which $F$ is injective. Thus, $F(u,v)=(0,0)$ has a unique solution in this neighbourhood. Hence, $(x,y)$ (from before) was the only solution. Since each point in this set is isolated then any cluster point must not belong to this set. – 2017-01-04
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    Putting this information together we obtain that there are no cluster points. – 2017-01-04
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    what I get is $ F = (F_1,F_2)$ is one one in some neighbourhood of (x,y) how are you talking about $F(u,v) = (0,0)$ and how the point $(x,y)$ is the solution? Sorry but can you elaborate or answer this? – 2017-01-04
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    I want to consider the map $(x,y)\to(f(x)\cos(y)-a,g(x)\sin(y)-b)$ which I will now denote by $F$. Note that $F(x,y)=(0,0)$ if and only if $f(x)\cos(y)=a$ and $g(x)\sin(y)=b$. Thus, for fixed $a,b$ we have: $$E_{a,b}=\left\{(x,y)\in\mathbb{R}^{2}\bigm\vert f(x)\cos(y)=a,\,g(x)\sin(y)=b\right\}=F^{-1}(0,0)$$ using your proof, we found that each point of $E_{a,b}$ has a neighbourhood, $U$, in which $F$ is injective. Thus, $F(u,v)=(0,0)$ can have at most one solution in $U$. This separates one point of $E_{a,b}$ from the rest. – 2017-01-04
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    Thanks. That solved my problem – 2017-01-04
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    You're welcome. Glad to help. – 2017-01-04

0 Answers 0