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Shafarevich in the book "Basic Algebraic Geometry I" gives the following definition of a quasi-projective variety:

A quasi-projective variety is an open subset (respect to the induced Zariski topology) of a closed projective set.

Then he says that a closed affine set is a quasi-projective variety, but i disagree with this statement. Here I'll try to show why:

We know that $\mathbb P^n_k=U_0\cup\ldots\cup U_n$, where $U_i$ is open and $U_i\cong\mathbb A_k^n$. If $X$ is a closed affine set and $\overline X$ is its projective closure (note that $\overline X$ is a closed subset of $\mathbb P^n_k$ and it contains a "copy" of $X$), one can show that $X$ is homeomorphic to $\overline X\cap U_i$ which is open in $\overline X$. So technically $X$ is homeomorphic to a quasi-projective variety, but according to the previous definition it is not a quasi-projective variety because $X\not\subset\mathbb P^n_k$. Where is the mistake in my argumentation?

Clearly Shafarevich is working into the classical framework of Algebraic Geometry, so without mentioning the concept of scheme or sheaf. For this reason I'd like an answer concerning only "classical arguments". I know that this is not the most elegant way to introduce varieties, but I want to understand the abstract concepts through sucessive generalizations.

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For the argument see If I understand correctly, then you should consider $X$ as embedded in projective space. – Dietrich Burde Oct 7 '13 at 18:44
This is the point! Considering $X$ embedded in $\mathbb P^n_k$ formally is not the same thing of writing $X\subseteq\mathbb P^n_k$. Maybe one should change the definition of quasi-projective variety a little bit. – Dubious Oct 7 '13 at 18:50
It depends on what your definition of "is" is. – Andrew Oct 7 '13 at 18:57
The definition of "quasi-projective" should be "isomorphic to an open subvariety of a closed subvariety of a projective space". It is a bad idea to impose conditions of equality on objects of a category. In General, all properties should be invariant under isomorphism. – KotelKanim Oct 7 '13 at 20:11
@Andrew I'd have a hard time giving a non-circular definition of "is"! – Bruno Joyal Oct 7 '13 at 21:25

That's because you have to think of $\mathbb A^n_k$ as being embedded in $\mathbb P^n_k$. The idea is that you should identify $\mathbb A^n_k$ with one of the open chart of $\mathbb P^n_k$. In this way an affine varieties in $\mathbb A^n_k$ become a quasi-projective variety (because it's a closed in $\mathbb A^n_k$ and so the intersection of a closed set in $\mathbb P^n_k$ with the open set $\mathbb A^n_k$).

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Vai Giorgiomossa! Sei tutti noi! – Joseph Curwen Oct 7 '13 at 22:52

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