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Here's my question. Let's consider the polynomial $p(y_{1},...,y_{n})$ with $deg(p)=d$. The set $C=\{(y_{1},...,y_{1})\in \mathbb{R}^{n} | p(y_{1},...,y_{n})=0\}$ is closed in $\mathbb{R}^{n}$ (because is the inverse image of the closed set $\{0\}$ in $\mathbb{R}$ under the function $p:\mathbb{R}^{n}\to \mathbb{R}$). When I want to consider this polynomial in the projective space $\mathbb{P}^{n}(\mathbb{R})$ I calculate the projective closure of it and so i consider the polynomial $P([x_{0},...,x_{n}])={x_{0}}^d p(\frac{x_{1}}{x_{0}},...,\frac{x_{n}}{x_{0}})$. My question is: Why is the zero set $C'=\{[x_{0},...,x_{n}]\in \mathbb{P}^{n}(\mathbb{R}) | P([x_{0},...,x_{n}])=0\}$ closed ? This time I can't use the inverse image of $P$ cause $P$ is not even a function ($P([x_{0},...,x_{n}])\neq P([\lambda x_{0},...,\lambda x_{n}])$)!

P.S= I'm considering the Euclidean topology in $\mathbb{R}^{n}$ and the final topology in $\mathbb{P}^{n}(\mathbb{R})$ (the finest topology such that $\pi :\mathbb{R}^{n+1}-{0}\to \mathbb{P}^{n}(\mathbb{R})$is continuous).

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But you can look at the zero set $D$ of $P$ in $\mathbb R^{n + 1}$. If $D = \pi^{-1}(C')$, then since the topology on $\mathbb P^n$ is the quotient topology induced by $\pi$ you are done. Is this the case? –  Dylan Moreland Feb 2 '12 at 16:42
That was soooo easy and clear.. I'm sorry I asked such a stupid question, but I'm really near my exam and I'm freaking out.. thanks Dylan for your help. –  Lorenzo Rossi Feb 2 '12 at 16:54
No need to apologize. Good luck! –  Dylan Moreland Feb 2 '12 at 18:38

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