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Is

$$\begin{align*} \left\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:(x_{1}x_{3}+x_{2}x_{4})\sin^{2}\theta+\left(x_{1}x_{4}+x_{2}x_{3}\right)\cos^{2}\theta\ne x_{1}x_{2}+x_{3}x_{4}~\forall\theta\in\mathbb{R}\right\} \end{align*}$$

a dense subset of the Euclidean $4$-dimensional space in topology? Thanks in advance for any help.

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What do you think, and why? –  Chris Gerig Jan 4 '13 at 21:04
    
I took the liberty of modifying the description of the set slightly so as to make it fit on one line. –  Brian M. Scott Jan 4 '13 at 21:16

1 Answer 1

No. Let $x_2=x_1+r_1,x_3=x_1+r_2$ and $x_4=x_1+r_3$. In order for $(x_1,x_2,x_3,x_4)$ to be in the set, we must have $$(2x_1^2+x_1(r_1+r_2+r_3)+r_1r_3)\sin^2\theta+(2x_1^2+x_1(r_1+r_2+r_3)+r_1r_2)\cos^2\theta\ne 2x_1^2+x_1(r_1+r_2+r_3)+r_2r_3$$ for all $\theta$, and since $\sin^2\theta+\cos^2\theta=1$ this simplifies to $r_1r_3\sin^2\theta+r_1r_2\cos^2\theta\ne r_2r_3$ for all $\theta$. But if $1<r_1<r_3$ and $r_2=1+\epsilon$ where $|\epsilon|$ is sufficiently small we get that $r_1r_3>r_2r_3>r_1r_2$ and since $r_1r_3\sin^2\theta+r_1r_2\cos^2\theta$ takes on both $r_1r_3$ and $r_1r_2$, by the IVT equality holds somewhere. Thus the set avoids a neighborhood of $(x_1,x_1+r_1,x_1+1,x_1+r_2)$ for any $x_1$ and $1<r_1<r_2$.

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Alex, thank you for your answer, but why "avoids a neighborhood of $(x_1,x_1+r_1,x_1,x_1+r_2)$ for any $x_1$ and $1<r_1<r_2$"? Didn't you set $x_3=x_1+r_2$? Thanks. –  sarira Jan 5 '13 at 1:01
    
Also, it seems that you constructed a stripe shaped neighborhood, but is the neighborhood 3-dimensional or 4-dimensional? Thanks. –  sarira Jan 5 '13 at 1:15
    
@sarira Your first comment is quite correct, I have edited my answer. To address your second, we indeed have a 4-dimensional neighborhood as we can vary $r_1$ and $r_2$ as long as they obey $1<r_1<r_2$ and we keep $\epsilon$ sufficiently small. And of course $x_1$ can vary arbitrarily. –  Alex Becker Jan 5 '13 at 4:17

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