Convexity of rint

Question

Let $K$ be a convex set in a real Hausdorff topological vector space. Recall that $\rm{rint}(K) = \{x \in K \;| \; \exists U \text{ an open neighborhood of } x : U \cap \rm{aff}(K) \subseteq K\}$. Where $\rm{aff}(K)$ denotes the affine hull of $K$.

Let $x,y \in \rm{rint}(K)$, then there exist $U_x$ and $U_y$ veryfing $U_x \cap \rm{aff}(K) \subseteq K$ and $U_y \cap \rm{aff}(K) \subseteq K$. Let $\alpha \in ]0,1[$ and $V:= \alpha U_x + (1-\alpha) U_y$, How to prove that $U \cap \rm{aff}(K) \subseteq K$?

Any ideas ?

Thank you for help.

Answer 1

It seems that without loss of generality we may suppose that $K$ contains the zero of the linear space and then the set $\mbox{rint}(K)$ is open as an interior of a convex subset of a linear space $\mbox{aff}(K)$.