Relative interior of the sum of two convex sets I'd like to show ri(C1-C2)=ri(C1)-ri(C2) without using the fact that relative interior is preserved under linear transformations. I.e. Is there a way to show this by showing both inclusions?
 A: Yes you can show it by both inclusions. It is tedious but good exercise.


*

*$ri(C_1-C_2)\subset ri(C_1)-ri(C_2)$. Proof: pick $c\in ri(C_1-C_2) \Leftrightarrow    c=\alpha x + (1-\alpha) y$ for $x,y\in C_1-C-2$ and $\alpha \in(0,1)$. Moreover $x\in C_1-C_2$ if and only if $x=x_1-x_2$ with $x_1\in C_1$ and $x_2\in C_2$ and the same goes for $y$. So we have $y=y_1-y_2$ with  $y_1\in C_1$ and $y_2\in C_2$. $\;$ Now : $$ c= \alpha x + (1-\alpha) y=\underbrace{\alpha x_1 + (1-\alpha) y_1}_{\in\, ri(C_1)} - \left(\underbrace{ \alpha x_2+(1-\alpha) y_2}_{\in\, ri(C_2)} \right)\in ri(C_1)-ri(C_2)$$

*$ri(C_1-C_2)\supset ri(C_1)-ri(C_2)$. We want to pick an arbitrary $c\in ri(C_1)-ri(C_2)$ and show it belongs to $ri(C_1-C_2)$. It is essentially doing step 1 backwards. I will leave this as an exercise.

*Above assumes the space is finite dimensional so the algebraic interior coincides with the relative interior for convex sets. To prove the first "iff": if x is in the algebraic interior of a convex set then for any y and for some t (scalar) we have x+t.y also is in the algebraic interior, call this point x1=x+t.y. Now consider the point −y then there is some also some small scalar r such that x2=x+r(−y) is also in the algebraic interior. Clearly x lies in the line that connects x1 and x2 so then we can express x as a convex combination them.  

*Jan van Tiel's little book: "Convex Analysis, An Introductory Text" is a wonderful help for these type of problems.

A: See Corollary 6.6.2 (pp. 49) of "Convex analysis" by Rockafelar.
A: There is actually a simple intuitive explanation.


*

*We first show the forward inclusion must be true, i.e., $ri(C_1 - C_2) \subset ri(C_1) - ri(C_2)$.


Suppose $x\in ri(C_1 - C_2)$, then $\exists, y\in C_1, z\in C_2$ such that $x = y-z$. There are three cases: 
a) If $y \in ri(C_1)$ and $z\in ri(C_2)$, then the forward inclusion is true.
b) If $y \in ri(C_1)$ and $z\notin ri(C_2)$, then we can tweak $y$ and $z$ to be $y'$ and $z'$ such that $y' \in ri(C_1)$ and $z' \in ri(C_2)$ and $x = y'-z'$.
c) If $y \notin ri(C_1)$ and $z\notin ri(C_2)$, then $x=y-z \notin ri(C_1 - C_2)$


*We now show the backward inclusion must be true. It is true because of part a) in the forward inclusion.


These are intuitive explanation. You need to translate them in mathematical language, but it should be easy to write out the details.
