The Zariski closure of $X$ coincides with $(\mathbb{P}^{g-1})^*$ What does it mean to say: " the Zariski closure of $X$ coincides with $(\mathbb{P}^{g-1})^*$ " ? 
would be: $\overline{X}=(\mathbb{P}^{g-1})^* $ ? Is correct? If it is correct, then, in general, to justify I need to show that $ X $ is dense in $(\mathbb{P}^{g-1})^* $? 
Thanks in advance!
 A: The Zariski closure means the closure with respect to the Zariski topology, which roughly speaking is defined by taking as closed sets the common zero loci of a family of polynomials (homogeneous polynomials in the case of projective sapce).
Assuming that your upper star means dual, then if $\mathbb{P}=\mathbb{P}(V)$ for some vector space $V$ (e.g. $\mathbb{C}^{g}$), its dual is given by $\mathbb{P}(V^{*})$. In particular, it is also irreducible. So the Zariski closure of any nonempty open susbset of $(\mathbb{P}^{g-1})^{*}$ is the whole space $(\mathbb{P}^{g-1})^{*}$. So you can check that $X\neq \varnothing$ and that there is a Zariski open set contained in $X$. For example, if the complement of $X$ is contained in the zero locus of a polynomial.
It can also happen that $X$ is dense without containing any open set, for example a proper subset of infinitely many points of a curve is dense in the curve, and so on.
Projective $n$-space is covered by $n+1$ copies of affine $n$-space, which are Zariski open. You can check that the intersection of $X$ with each of those is dense in the corresponding affine space, which is probably easier to check (you can drop the homogeneity restriction on polynomials).
I don't think there is a general characterization for Zariski dense sets. I guess you have to use something like these things said above to prove it in your particular case.
