Intuition behind , and understanding the definition of convexity in the real projective space I have been studying the Real projective space, which we denote by $\mathbb{RP}^n$, is obtained by identifying any two points $\mathbb{R}^{n+1} \setminus \{0\}$ that are proportional. I have also understood how in $\mathbb{RP}^2$, projective points are lines through the origin, and projective lines are planes containing the origin.
I am reading the book Complex Convexity and Analytic Functionals by Andersson .
Here definition 1.3.2 says A subset $E$ of $\mathbb{RP}^n$ is said to be convex if it does not contain any projective line and its intersection with any projective line is connected. A convex set is called non-degenerate if it is not contained in any hyperplane and does not contain any affine line.
Can anyone tell me first as to how I can relate this definition to convexity in the Euclidean space? This way I can get an intuition behind the above definition. Why is a convex set not supposed to contain a projective line? Why do we consider and define non-degenerate convex set?
 A: First I'd like to point out that I don't study convexity or anything even close to it (I study algebraic geometry), so take what I have to say with a grain of salt.

Can anyone tell me first as to how I can relate this definition to convexity in the Euclidean space?

A set $A\subset \mathbb R^n$ is convex if given any two points $x,y\in A$, the line segement $\overline{xy}$ with endpoints $x$ and $y$ is contained in $A$. $A$ is line convex (a term I just made up) if $A\cap \ell$ is connected for every line $\ell$. I'll leave it as an exercise to show that these notions are equivalent, that is, $A$ is convex if and only if $A$ is line convex. The second definition translates better to projective space however, because for any two points $x,y\in \mathbb{RP}^n$ there are two line segments joining $x$ and $y$. If we used the first definition then some choice would be required. (I imagine that this choice would lead to sets being given the title of convex without seeming like they should be convex, but I can't come up with one off the top of my head. I'll update this answer if I think of one.)

Why is a convex set not supposed to contain a projective line?

If you allow $A\subset \mathbb{RP}^n$ to contain a line, then you necessarily have that lines are convex. (Real) projective lines are circles, which I don't think should be convex (they have a hole). However, (real) projective lines with a point deleted are just lines, so requiring that entire lines are not contained in $A$ seems fitting. (I can't really give a better answer than this here. I imagine that there are some nice properties that you want convex sets to have in general that circles don't have [contractibility maybe?]. Your book might have some of these results. If you find some reason then I'd love to hear them.)

Why do we consider and define non-degenerate convex set?

This sort of thing, requiring that something not be contained in any hyperplane, shows up a lot in the study of projective space. My understanding of why this is made explicit a lot of the time is because something contained in a hyperplane is contained in some smaller $\mathbb{RP}^m$. For me, I would care because I can only guarantee that things of certain codimensions intersect in nice ways, from which I can get more information. For you I imagine it has to do with bounding problems. A common problem when studying convexity (I believe) is to give a certain list of requirements and then ask questions like "what is the convex set of least volume which satisfies these requirements" (cf. the Kakeya needle problem). For these you want to talk about non-degnerate convex sets because  subsets of hyperplanes have zero volume, and so this would trivialize these problems in an undesirable way.
A: Definition : $A\subset \mathbb{S}^n$ is convex if for any $x,\ y\in
A$, the any shortest path $[xy]$ is in $A$.
If $\pi : \mathbb{S}^n\rightarrow
  \mathbb{R}P^n
$ is quotient map i.e. $\pi (x)=\pi(-x)$, then consider a set
$E\subset \mathbb{R}P^n$ s.t. $\pi^{-1}E=E_1\bigcup E_2 $ where
$E_1$ is an antipodal image of $E_2$ and a closed set $E_1$ is in a
some open hemisphere.
Further assume that $E_1$ is convex. When $C$ is a great circle in
$\mathbb{S}^n$, then $\pi^{-1} E$ does not contain $C$ so that $E$
does not contain any projective line.
Further note that $E_1 \bigcap C$ is connected so that an
intersection between $E$ and any projective line is connected.

Hence
  any compact convex set in open hemisphere defines a convex set in $
\mathbb{R}P^n$

