I 'm new to algebraic geometry and I am a little bit confused with a proposition on Fulton, Algebraic Curves, page 70 proposition 3: I will quote the proposition:
Let V be a closed subvariety of $P^n$, $\phi_i : A^n \to U_i \subset P^n$, (whereas $\phi_{n+1}(x_1,...,x_n)=(x_1:...:x_n:1)$ ). Then $V_i=\phi_i^{-1}(V)$ is a closed subvariety of $A^n$, and $\phi_i$ restricts to an isomorphism of $V_i$ with $V\cap U_i$. A projective variety is a union of a finite number of open affine varieties.
It seems straightforward to show that $V_i=(V\cap U_i)_\ast$ and thus prove that it is a closed subset of $A^n$ and with a little bit of extra effort prove that it 's also irreducible. Skipping the isomorphism part, I am confused with the last statement that a projective variety is a finite union of open affine varieties (the finite part is easy to show I guess from Hilbert's Basis Theorem, what I m confused is that they 're open). From the above, $V_i$ is closed, not open.
In general, I have seen the reverse statement going like this: Let $X=V(f)$ be an affine variety of $A^n$ and $D(f)=\{P \in A^n: f(P) \neq 0\}$. Obviously $D(f)$ is open. Then consider the map $\phi: X \to A^{n+1}$, where $\phi(x_1,...,x_n)=(x_1,...,x_n,1/f(x_1,..,x_n) )$. Then you can indeed prove that $\phi(X)=V(fX_{n+1}-1)$ and thus a closed subvariety of $A^{n+1}$. It seems to me that you can somehow reverse engineer this logic for the above proposition but I am not sure how.
If anyone had the patience to help me understand it, consider the following trivial example: Let $n=1, A^1, P^1$ and $V=V(X-Y)=\{(x,x), x\in A^1\}$. If $f=X-Y$, then $f_\ast = X-1$, and $V_2=V_\ast =V(X-1)=\{1\}$. How can you show in this example that V is a union of open affine varieties?
I 'm sorry in advance if I have misunderstood anything, any help would be appreciated.