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I try once more: Consider the two sets $C_1$ and $C_2$ that are defined as follows: $$ \left\{ \begin{array}{ll} C_1=\{\,(x_1,x_2,x_3)\in \Bbb{R}^3 \mid x_3\ge 0\,,\,x_3^2 \ge x_1^2+x_2^2\,\} &,\\ \\ C_2=\{\,y\in \Bbb{R}^3 \mid \forall x \in C_1 \,,\, y^t\cdot x \le 0\,\} &. \end{array} \right. $$ Suppose that $$ C_3 =\{\,y\in \Bbb{R}^3 \mid \forall x \in C_1\cap \Bbb{Z}^3 \,,\,y^t\cdot x \le 0 \, \} $$

My question: How to prove that $C_2=C_3$?

Thanks for any suggestions.

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  • $\begingroup$ I'm not sure that $C_2=C_3$, it looks like $C_3$ dense in $C_2$ $\endgroup$ – Matheman Jun 8 '17 at 8:14
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Note that it is not hard to show that for any $x \in C_1$ there exist $z \in C_1 \cap Z^3$ such that $\|x-z\| \leq \sqrt3$. Now taking into account the later statement we are ready to prove your claim.

Proof: first since $ C_1 \cap Z^3 \subset C_1 $ then we get $C_2 \subseteq C_3.$ For the other direction take $y \in C_3$ we want to show that $\langle y,x\rangle \leq 0 $ for all $x \in C_1$. To this ends, suppose there exists $x \in C_1$ such that $\langle y,x \rangle > 0.$ Now for all $n \in N$ there exist $z_n \in C_1 \cap Z^3$ such that $\|nx-z_n\| \leq \sqrt 3. $ Therefore for all $n \in N$ we have

$$ n \langle y ,x \rangle - \langle y,z_n\rangle = \langle y, nx-z_n \rangle \leq \sqrt 3 \|y\| $$ Now due to our choice of $x,z_n \in C_1$ left side converges to $+ \infty$ as $n \rightarrow + \infty.$ which is a contradiction.

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