Given the projective variety $X=\{w_o^2+w_1^2+w_2^2+w_3^2=0\}\subseteq \mathbb{P}^3$, we define $X_1=X-\{(1,i,0,0)\}$. Is $X_1$ projective/affine/neither?
I have tried to prove that is is affine, by constructing a map $f: X \to \{(x-1)(y-i)z\cdot t=1\} \subseteq \mathbb{C}^4$ by $f(w_0,w_1,w_2,w_3)=(w_0,w_1,w_2,\frac{1}{(w_0-1)(w_1-i)w_2})$ (just like in the affine case). Is this legit? I mean, this is not even well defined.
How to solve this question in general?
a) Here is an amazing, too little known general theorem of the type you request:
If $X$ is a completely arbitrary variety and $U\subset X$ is an affine open subset, then the codimension of $X\setminus U$ in $X$ is $1$.
This immediately implies that $X_1$ is not affine.
b) On the other hand a subset of a projective variety is projective if and only if it is closed, so that $X_1$ is not projective either.
Edit
$\bullet$ The theorem alluded to in a) is due to Goodman and can be found as Proposition 1, page 162 in Annals of Mathematics, Second Series, Vol. 89, No. 1 (Jan., 1969)
$\bullet \bullet$ The non-trivial implication of result b) follows from the fact that a morphism from a projective variety to an arbitrary variety (here the open immersion inclusion $X_1\hookrightarrow X$) has closed image.
A proof can be found in this handout by Patrick Morandi