TITLE: Is convex open set in $\mathbb{R}^n$ is regular?
QUESTION [5 upvotes]: Convex set $A$ in $\mathbb{R}^n$ iff for every $ x,y\in A,0\le\lambda\le1$, then $\lambda x+(1-\lambda)y\in A$.
$U$ is regular open set iff $A$ is open and $int(cl(A)) = A$.
Is convex open set in $\mathbb{R}^n$ is regular?

REPLY [9 votes]: I see the situation as follows. Let $A$ be a nonempty open convex subset of $\mathbb{R}^n$. Fix an arbitrary point $x_0\in A$. Since $A$ is open, there exists a small ball $B$ such that $x_0\in int(B)\subset A$. Suppose that there exists a point $x\in int(cl(A))\backslash A$. Consider a ray $r$, which is going from the point $x_0$ to the point $x$. Since $x\in int(cl(A))$, there there is a point $x_1\in cl(A)$, such that $x_1$ is laying behind $x$. Thereore there is a number $\mu>1$ such that $x_1+\mu(x-x_1)=x_0$. Since $x_0\in int(B)$ and the addition and the scalar multiplication are continuous, there is a neighborhood $U$ of $x_1$ such that $b=a+\mu(x-a)\in B$ for each $a\in U$. Hence $x=(1/\mu)b+(1-1/\mu)a\in conv(B\cup \{a\})$, where $conv(B\cup \{a\})$ is the convex hull of the union $ B\cup \{a\}$. Since $x_1\in cl(A)$, there a point $a\in A\cap U$. Since the set $A$ is convex, $A\supset conv(B\cup\{a\})\ni x$, a contradiction.