Is convex open set in $\mathbb{R}^n$ is regular?

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?

Solution 1:

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.

Solution 2:

Yes. Let $A\subseteq \mathbb R^n$ be an open convex set such that the origin is in the interior of the closure of $A$.

This means there exists $t>0$ such that the open box $$\{x\in\mathbb R^n\mid -t\leq x_1,\dots,x_n\leq t\}$$ is a subset of the closure of $A$.

This implies that $A$ has a point in each orthant of $\mathbb R^n$: for all $p_1,\dots,p_n\in\{-,+\}$, there exists $x(p_1,\dots,p_n)\in A$ such that for each $1\leq i\leq n$, the sign of $x_i$ is $p_i$. For each $p_1,\dots,p_{n-1}\in\{-,+\}$, define $$x(p_1,\dots,p_{n-1},0) = \frac {x(p_1,\dots,p_{n-1},-)x(p_1,\dots,p_{n-1},+)_n+x(p_1,\dots,p_{n-1},+)x(p_1,\dots,p_{n-1},-)_n} {x(p_1,\dots,p_{n-1},+)_n+x(p_1,\dots,p_{n-1},-)_n}.$$ So the sign of $x(p_1,\dots,p_{n-1},0)_i$ is $p_i$ for $1\leq i\leq n-1$, while $x(p_1,\dots,p_{n-1},0)_n=0$. Note that this is a convex combination. So $A$ has a point in each orthant of the $\{x\in\mathbb R^n \mid x_n=0\}$ hyperplane. Continuing in this way (or by induction on $n$) we find that $A$ contains $0$.