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Is the circle compact in $\mathbb{P}_{2}(\mathbb{C})$?

Here what I did: I considered the circle in $\mathbb{C}^2$ is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$. The projective closure in $\mathbb{P}_{2}(\mathbb{C})$ is $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$. The points at infinity are $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}\cap\{{x}_{0}=0\}=\{[0,1,i],[0,1,-i]\}$. So for every open cover ${\{{A}_{i}\}}_{i\in I}$ there will be a $j$ and a $k$ in $I$ fow which $[0,1,i]\in{A}_{j}$ and $[0,1,-i]\in{A}_{k}$. Therefore ${\{{A}_{i}\}}_{i\in I-\{j,k\}}$ is an open cover of the affine part $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$ and (if what i wrote is correct) the circle in $\mathbb{P}_{2}(\mathbb{C})$ is compact iff the circle $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ is compact in $\mathbb{C}^2$. A subspace is compact in $\mathbb{C}^2$ iff it's close and limitated. Considering the continuous function $f:\mathbb{C}^2\to\mathbb{C}$ defined by $(x,y)\mapsto x^2+y^2$, the circle is ${f}^{-1}(1)$ so it's closed ($\mathbb{C}$ is T1, the points are closed). But it's limitated? Seems to me it is not.

Can anyone help me. Is what i wrote correct? Is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ limitated in $\mathbb{C}^2$?

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    Thank you very much... really usefull. So my reasoning is correct (with the correction Aaron made). Thanks again for your time and disponibility. P.S : And yoyo sorry for the word limitated, I've wrote the question quickly and I've wrongly translated the word from Italian.2012-01-27

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