Prove that convex open set in $\mathcal{R}^d$ are algebraically open.

Question

I am reading Barvinok's A First Course in Convexity. I am not sure how to write the intersection of any convex set in $\mathcal{R}^d$ and straight line as the form of open interval. I do not have backgroud in topology. Any help is appreciated.

Answer 1

The usual definition of "algebraically open" is:

A set $V \subset \mathbb R^d$ is algebraically open if, for each $u,v \in \mathbb R^d$ the set $V_{u,v}:=\{ t \in \mathbb R : u+tv \in V \}$ is open.

It is easy to prove that each open set is algebraically open (convexity is not needed). Indeed, if $t \in V_{u,v}$ then $u+tv \in V$. Since $V$ is open, there exists some $r>0$ so that $B_r(u+tv) \subset V$.

Now, it is easy to see that $$|t-s|<\frac{r}{\| v \|+1} \Rightarrow \| (u+tv)-(u+sv)\|=|t-s| \|v \| < r \Rightarrow u+sv \in B_r(u+tv) \subset V \Rightarrow s \in V_{u,v}$$

This shows that whenever $t \in V_{u,v}$ there exists some $r>0$ such that $(t-\frac{r}{\| v \|+1} , t+\frac{r}{\| v \|+1} ) \subset V_{u,v}$.

Answer 2

Suppose $C$ is convex and $L$ is a line. In particular, we can write $L = \phi(\mathbb{R})$, where $\phi(t) = x_0+td$.

Let $I = \{ t | \phi(t) \in C \}$. It is straightforward to check that $I$ is a convex subset of $\mathbb{R}$.

The intervals are exactly the convex subsets of $\mathbb{R}$.

Note that $I$ is not necessarily open, for example, if we take $C$ to be the set $\{x \in \mathbb{R}^b | |x_k| \le 1 \}$, $x_0 = 0$, $d = e_1$ then $I=[-1,1]$

Regarding notation, if $a,b \in \mathbb{R}^n$ (or any vector space) then we write $[a,b]$ to be the convex hull of $a,b$, and the obvious alterations excluding one or other end point is written $[a,b)$, etc. Of course, the sets $[a,b]$ and $[b,a]$ are the same using this interpretation.