If $A$ and $B$ don't contain rays, can $Conv(A\cup B)$ contain a ray?

Question

Let $X$ be a normed space and let $A,B\subseteq X$ be non-empty closed convex sets with $0\in A\cap B$ such that they don't contain rays of the form $\{tx:t\ge 0\}$ (for some $x\in X$).

Can we assure that $Conv(A\cup B)$ doesn't contain any ray of the form $\{tx:t\ge 0\}$ for some $x\in X$?

I thought this is true but now I'm not sure. I tried to prove that if $Conv(A\cup B)$ contains a ray then so do $A$ and $B$, but I couldn't do this.

Any hints?

Edited: I added that $A,B$ are closed.

Answer 1

In finite dimensions, it's sufficient to assume $A$ and $B$ are either open or closed; indeed we have

If $X$ is finite dimensional, $A$ is convex, contains $0$, contains no ray, and is unbounded, then $A$ is neither closed nor open.

Indeed pick $a_n\in A$ with $|a_n|\geq n$. Because the unit sphere is compact, by passing to a subsequence we may suppose $a_n/|a_n|\to a$ as $n\to\infty$. Since $A$ contains no ray, there exists $t>0$ with $ta\notin A$. For $n>t$ we have $|a_n|>t$, so $ta_n/|a_n|\in A$. This converges to $ta$, so $A$ can't be closed. If $A$ is open then $B(0,r)\subseteq A$ for some $r>0$. Then $A$ contains $\frac{r(|a_n|a-a_n)}{2||a_n|a-a_n|}$, so it contains $$ \frac{r(|a_n|a-a_n)+ra_n}{2||a_n|a-a_n|+r} =\left(\frac{r|a_n|}{2\left||a_n|a-a_n\right|+r}\right)a. $$ Thus $$ 2\left|a-a_n/|a_n|\right|+r/|a_n|>r/t, $$ giving a contradiction as $n\to\infty$.

Now consider the following subset of $\ell_1$: $$ A=\left\{x\in\ell_1\mid x_i\in[0,1]\text{ for }i>0\text{ and }\|x\|_1=2x_0\right\}. $$ It is easy to see that $A$ is closed and convex. Let $\mu:\ell_1\to\ell_1$ be the linear map which switches the sign of all but the first entry, and let $B=\mu(A)$. Fix $n$ and consider the sequence $x$ given by $x_0=n$, $x_i=1$ for $1\leq i\leq n$, $x_i=0$ for $i>n$. Then $x\in A$ so $\mu(x)\in B$, and $$ \frac12(x+\mu(x))=(n,0,0,\ldots). $$ Since $n$ was arbitrary, $\mathrm{conv}(A\cup B)$ contains the ray spanned by $(1,0,0,\ldots)$.

Finally without an open or closed assumption, the statement fails even in $\mathbb R^2$: $$ A=\{(x,y)\mid y\geq x^2,\,x>0\}\cup\{(0,0)\}, $$ $$ B=\{(x,y)\mid y\geq x^2,\,x<0\}\cup\{(0,0)\}. $$

Answer 2

If $\{tx\mid t\geq 0\} \subset {\rm conv}\ A\cup B$ with $x\neq 0$, then there is $x_i > 0 $.

$\pi_i:\prod_{i=1}^\infty\ \mathbb{R} \rightarrow \mathbb{R}$, then $$\pi_i\ {\rm conv}\ A\cup B\supseteq \{t\mid t\geq 0\} $$

Hence we can assume that $\pi_i\ A=\{t\mid t\geq 0\}$.

Hence $(0,\cdots, 0,\underbrace{t}_{\stackrel{i\textrm{-th}}{\text{ coordinate}}},0,\cdots ) \in A,\ t\geq 0 $ so that it is a contradiction.