Projective space definitions My questions are as follows:

  
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*Are all these different definitions of projective space equivalent? For example, Bezout's theorem holds under all 4 definitions (with an appropriate change in terminology)? 
  
*Are the projective topological spaces obtained under Zariski and usual topology homeomorphic? (though both are compact but the one with Zariski is not Hausdorff)
  



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*Algebraic geometry [Hartshorne]


Firstly, we define affine n-space over an algebraically closed field $k$, denoted as $\mathbb{A}^n$, to be the set of all n-tuples of elements over $k$. Then we give it Zariski topology where the closed sets are the vanishing of polynomials, i.e. for $T\subset k[x_1,\ldots, x_n]$ the closed sets are $Z(T)=\{p\in \mathbb{A}^n: f(p)=0 \ \forall f\in T\}$. 
Secondly, we define projective n-space over $k$, denoted as $\mathbb{P}^n$, to be the set of equivalence classes of $(n+1)$-tuples of elements over $k$, not all zero, under the equivalence relation given by $(a_0,\ldots,a_n)=(\lambda a_0,\ldots, \lambda a_n)$ for all $\lambda\in k\backslash {0}$. Then we give it Zariski topology where the closed sets are the vanishing sets of homogeneous polynomials, i.e. for $T\subset k[x_1,\ldots, x_n]$ set of homogeneous poylnomials (graded ring) the closed sets are $Z(T)=\{p\in \mathbb{P}^n: f(p)=0 \ \forall f\in T\}$. 
Then we observe that both of these spaces are Noetherian topological spaces, and hence they are compact (w.r.t. Zariski topology) but not Hausdorff (since infinite sets). Moreover, we can show that any two curves in $\mathbb{P}^2$ intersect, as well as powerful statements like Bezout's theorem (whihc we can use to prove Pascal's theorem about conics).


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*Algberaic topology [Hatcher]


The projective $n$-space $\mathbb{F}\mathbb{P}^n$ is the space of all lines through the origin in $\mathbb{F}^{n+1}$, where $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Each such line is determined by a nonzero vector in $\mathbb{F}^{n+1}$, unique up to scalar multiplication, and $\mathbb{F}\mathbb{P}^n$ is topologized as the quotient space of $\mathbb{F}^{n+1}-0$ under the equivalence relation
$v\sim \lambda v$ for scalars $\lambda \neq 0$. 
Then we observe that $\mathbb{F}\mathbb{P}^n$ is a compact Hausdorff space. But, $\mathbb{C}\mathbb{P}^n$ and $\mathbb{R}\mathbb{P}^n$ have different CW-complex decomposition and homology groups.


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*Projective Geometry [Coxeter]


Axioms for the development of two-dimensional projective geometry:


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*Any two distinct points are incident with just one line.

*Any two lines are incident with at least one point.

*There exist four points of which no three are collinear.

*The three diagonal points of a quadrangle are never collinear.

*If a projectivity leaves invariant each of three distinct points on a line, it leaves invariant every point on the line.
Then, we can have various examples of theorems in finite geometry, denoted by $PG(n,q)$ where $q=p^m$ for some prime $p$ and positive integer $m$. We have Desargues theorem.


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*Projective space of a vector space [O'Connor]


Given any vector space $V$ over a field $F$, we can form its associated projective space $\mathbb{P}(V)$ by using the construction $\mathbb{P}(V) = V - \{0\}/\sim$ where $\sim$ is the equivalence relation $u \sim v$ if $u = 
\lambda v$ for $u, v$ belongs $V - \{0\}$ and $\lambda$ belongs $F$. 
Examples are $\mathbb{P}(\mathbb{R}^n)$, $\mathbb{P}(\mathbb{C}^n)$, can take $F$ to be a finite field. We have Desargues theorem.
 A: *

*They are clearly not all equivalent. For example, you yourself state that the topological spaces obtained from some of these definitions are Hausdorff and those obtained from some others are not. Hausdorffness is an homeomorphism invariant of topological spaces, so the Algebraic Geometry and Algebraic Topology definitions cannot produce homeomorphic topological spaces. Additionally, it's easy to construct two different smooth curves in $\mathbb {RP}^2$ that do not intersect, so Bezout's theorem can't hold for arbitrary smooth curves in projective space. It's also possible to break Bezout's theorem over $\mathbb{RP}^2$ using things like the conic $y=x^2$ and $y=-1$. These two curves do not intersect, as their two points of intersection are complex!

*As described in the above paragraph, the answer to this question is no.

*I'd quibble with your definition of affine space and projective space- the correct version of these is the scheme definition so you have access to generic points, sheaves, etc.
There are relations between each of these definitions, though. Given a scheme $X$ defined over a field $F$ with a topology, we can equip $X(F)$, the $F$ points of $X$, with a topology inherited from $F$. When $X=\Bbb{P}^n$ and $F=\mathbb{R}$ or $\mathbb{C}$, this gives the analytic topology on $\mathbb{RP}^n$ or $\mathbb{CP}^n$, respectively. Every analytic manifold may then be regarded as a smooth manifold, and every smooth manifold has a unique analytic structure. So the AG and AT definitions are very close together in some sense, but they do not define exactly the same thing.
Projective space of a vector space is easily seen to be a generalization of the AG definition as long as you don't squint too hard at what topology is happening. Instead of taking the projectivization of $F^{n+1}$, we take projectivizations of arbitrary vector spaces. This can be nice because it allows us to avoid choosing isomorphisms between arbitrary vector spaces and $F^k$ for some ordinal $k$.
The Coxeter definition is one I am less familiar with, but should be able to be regarded as $\Bbb P^n(F)$ for some field $F$ (again, without looking too hard at the topology - to truly make the description precisely equivalent or inequivalent to the others, you'll have to be clearer about how you're looking to consider this object).
The point is that projective space is something you can consider from many different viewpoints, and while you're always somehow talking about the same object, the context you're talking about it in can change important aspects of the construction, and it's important that you're clear about what you really mean.
