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?

There is actually a simple intuitive explanation.


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*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.
