a question about interior and closure $C$ is an non-empty subset of $R^n$ which has the following property: if $x,y \in C$, then $(x+y)/2 \in C$.
Let $a,b\in R^n$ such that $a \in C$ and $b \in $ interior of $C$. Prove that $(a+b)/2 \in $ interior of $C$.
If $a \in $ closure of $C$ and $b \in $ interior of $C$, does it still follow that $(a+b)/2 \in $ interior of $C$?
where a,b,x,y are all vectors in R^n, i type them as letters because i dont know how to type vectors. sorry for inconvenience!
i have tried to find a e>0 such that the open ball centered at p=(a+b)/2 with radius e is contained in C, but i failed.
i also tried the proof by contradiction: suppose p does not belong to interior of C. still,i did not get something useful.
 A: Consider the map $f(x)=(x+a)/2.$ This is a homeomorphism of $\mathbb R^n$ onto $\mathbb R^n.$ Thus $f$ is an open map. Hence $f(\text {int }C)$ is an open set. But $f(\text {int }C)$ is the set of midpoints of segments from points of $\text {int }C$ to the point $a.$ Therefore $f(\text {int }C)\subset C.$  Thus $(b+a)/2 \in f(\text {int }C) \subset \text { int }C$ as desired.
A: Take $r>0$ such that $B(b,r)=\{y: \|y-b\|<r\}\subset C.$
(1). When $a\in C$:  For any $x\in B((a+b)/2,r/2)$  we have $$\|(2x-a)-b)\|=2\|x-(a+b)/2\|<r,$$ $$\text {so }\quad 2x-a\in B(b,r)\subset C,$$ so $2x-a\in C$ so $x=((2x-a)+a)/2\in C.$ Hence $B((a+b)/2,r/2)\subset C.$
(2). When $a\in \bar C$: Take $a'\in C$ with $\|a'-a\|<r/2.$ By (1) we have $B((a'+b)/2,r/2)\subset C.$ Now for any  $x\in B((a+b)/2, r/4)$ we have $$\|x-(a'+b)/2\|\leq \|x-(a+b)/2\|+\|(a+b)/2-(a'+b)/2\|=$$ $$=\|x-(a+b)/2\|+\|(a-a')/2\|<r/4 +r/4=r/2,$$  so $x\in B((a'+b)/2,r/2)\subset C$ so  $x\in C.$ So $B((a+b)/2,r/4)\subset C.$
For (1), note that $2x-a$ is the point on the  line thru $a$ and $x$ that is twice as far from $a$ as $x$ is from $a.$ A diagram could clarify the  idea.
