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Can anyone tell me please that set of all symmetric non-negative definite matrices are closed or not in $\mathbb{M}_n(\mathbb{R})$ with usual topology

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  • If $S$ is a matrix and $x\in\Bbb R^n$, the map $f_x\colon S\mapsto x^tSx$ is linear hence continuous.
  • Let $S^+_n(\Bbb R)$ the set of symmetric non-negative definite matrices. Then $$S^+_n(\Bbb R)=\bigcap_{x\in\Bbb R^n}f_x^{-1}([0,+\infty)).$$
  • As $f_x^{-1}([0,+\infty))$ is closed in $\mathcal M_n(\Bbb R)$, $S_n^+(\Bbb R)$ is closed as an arbitrary intersection of such sets.
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I am confused by the answer of you and the link given by panu.please guide me properly – khuku Jan 20 '13 at 14:39
So, what don't you understand? – Davide Giraudo Jan 20 '13 at 14:39
they give different answer.there says that it is not closed – khuku Jan 20 '13 at 14:43
Positive definite is not the same thing as non-negative definite (which means $x^tSx\geqslant 0$ for all $x$). – Davide Giraudo Jan 20 '13 at 14:48
I can't understand how $S^+_n(\Bbb R)=\bigcap_{x\in\Bbb R^n}f_x^{-1}([0,+\infty)).$? Please explain. – S.Panja-1729 Feb 14 '15 at 19:08

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