There is a ray from each point of unbounded convex set that is inside the set. [closed]

Let $A$ be a non-empty convex, unbounded set in $\mathbb R^n$. Prove that for each point $a \in A$, there is a non-zero vector $h \in \mathbb R^n$ such that $l = \{x \in \mathbb R^n \mid x=a+th,\ t\ge0 \} \subset A$.

Solution 1:

The claim is incorrect. Consider the following counter-example: $$A=\left\{ \left(x,y\right)\in\mathbb{R}^{2}:0<x<1\right\} \cup\left\{ \left(0,0\right)\right\}\\a=\left(0,0\right)$$ It is correct if you require $A$ to be closed or $a$ to be in $\text{rel int}A$. Since for every non-empty convex set we have $\text{rel int}A\ne\emptyset$, it implies that every unbounded convex set contains a ray (but not necessarily from each point).

Solution 2:

For $r>0$, let $$V_r=\{h\in \mathbf S^{n-1}\mid a+rh\notin A\}.$$ By convexity, $r<s$ implies $V_r\subseteq V_s$. Let $$ U_t=\bigcup_{0<r<t}V_r.$$ Then $U_t$ is open, see below.

Assume $S^{n-1}\subseteq\bigcup_{t>0}U_t$, then by compactness of $S^{n-1}$ a finite subcover suffices and in fact $S^{n-1}\subseteq U_t$ for a single (maximal among finitely many) $t$. As $A$ is unbounded, there is $b\in A$ with $|b-a|>t$. With $h=\frac{b-a}{|b-a|}$, we have $h\in U_t\subseteq V_t\subseteq V_{|b-a|}$ hence $b=a+|b-a|h\notin A$, contradiction. Therefore, there exists some $h\in S^{n-1}$ that is not in any $U_t$ and also not in any $V_t$, hence $a+th\in A$ for all $t>0$ (and trivially also for $t=0$). $_\square$

Why is $U_t$ open? Assume $h\in U_t$ and specifically $h\in V_r$ with $0<r<t$. By the convexity of $A$, there exists $k\in\mathbb R^n$ with $\langle (a+rh)-x,k\rangle >0$ for all $x\in A$. Especially, $\langle h,k\rangle>0$ when $x=a$ because $a\in A$. Select $s$ with $r<s<t$. Then $(s-r)\langle h,k\rangle>0$ and for all $h'$ sufficiently close to $h$, we have $(s-r)\langle h,k\rangle +s\langle h'-h,k\rangle>0$. Hence for such $h'$ and all $x\in A$ $$\begin{align}\langle (a+sh')-x,k\rangle &=\langle (a+rh)-x,k\rangle+(s-r)\langle h,k\rangle +s\langle h'-h,k\rangle >0,\end{align}$$ especially $h'\in V_s\subseteq U_t$ as was to be shown.

Solution 3:

Of course if A is closed the result holds.

In fact if $x_0\in A$, then for each natural number n, it is possible to chose a sequence $l_n$ of line segments of length greater then $n$, with one endpoint $x_0$. If the sequence of points on $l_n$ which lies on the unit sphere around $x_0$ converges to $x$, then the ray staring at $x_0$ passing through $x$ is entirely inside A.

What if A is not closed ? Can we have at least one point in $A$ from where we can have an infinite ray contained inside $A$