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Prove that $\left\{ ( \sin a \cos b , \cos a\cos b , \sin b)\mid a , b \in \mathbb{N} \right\}$ is dense in the unit sphere.

Any help will be appreciated.

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    What have you tried? Is it the case that when $a$ and $b$ are real numbers, all points of the sphere are covered?2012-09-03
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    Yes,they are basically the spherical coordinates2012-09-03
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    @Ricky: consider your last remark: represent any point by spherical coordinates this way, only with $a$ and $b$ real numbers. Can you use density of $\mathbb{Q}$ to first show density of this set with $\mathbb{N}$ replaced with $\mathbb{Q}$? If so, is there a transformation you could use (with perhaps a trigonometric identity) to start with $a,b\in \mathbb{Q}$ and end with $a',b'\in \mathbb{N}$ that gives the same spherical-coordinate point?2012-09-03
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    Because $\pi$ is irrational, the points $(\cos(n), \sin(n))$ for $n \in \mathbb N$ are dense in the unit circle. And the parametrization $(\sin(a) \cos(b), \cos(a) \cos(b), \sin(b))$ provides a continuous map of the product of two circles onto the sphere.2012-09-03
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    @Robert Can you just elaborate a bit?2012-09-04
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    He's essentially saying that the map you have is a stereographic projection: http://en.wikipedia.org/wiki/Stereographic_projection Now combine that with the fact that the numbers are individually dense in $\mathbb{R}$ and therefore they get mapped dense in the sphere.2012-09-04
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    If $S_1$ is dense in $X_1$ and $S_2$ is dense in $X_2$, then $S_1 \times S_2$ is dense in $X_1 \times X_2$. If $f$ is a continuous map from $X$ onto $Y$ and $S$ is dense in $X$, then $f(S)$ is dense in $Y$.2012-09-04

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Because $\cos(n)$ and $\sin(n)$ (for n an integer) are dense on the interval $[0,1]$ (because $\pi$ is irrational) and those map continu to your coordinates.

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    Sorry i did not use tex. Is that the reason for downvote ?2012-09-04
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    I suspect the reason you got downvotes is that that you basically just copied Robert Israel's comment from above (without even saying so).2012-09-04