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?
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.