Show that $S_1\oplus S_2$ is a closed convex set.

Question

Let $S_1=\{\lambda \mathbf d_1:\lambda\ge 0\}$ and $S_2=\{\lambda \mathbf d_2:\lambda\ge 0\},$ where $\mathbf d_1,\mathbf d_2$ are non-zero vectors in $\mathbb R^n.$ Show that $S_1\oplus S_2$ is a closed convex set.

The idea that I have to prove it is to first show that $S_1$ is convex and then to prove that the direct sum of convex sets is convex.

So let $x,y\in S_1,t\in(0,1).$ We need to show that $tx+(1-t)y\in S_1$.

$tx+(1-t)y=tx+y-ty=t(x-y)+y=tz+y,$ from here can I say that it is in $S_1?$

$(y\in S_1,z=x-y\in \mathbb R^n, t\ge0)$

Answer 1

Hint

Let $\mathbf{x}, \mathbf{y} \in S_1$, then there exists $\alpha, \beta \geq 0$ such that $\mathbf{x}=\alpha \mathbf{d_1}$ and $\mathbf{y}=\beta \mathbf{d_1}$. Now consider $$t\mathbf{x}+(1-t)\mathbf{y}=\underbrace{[t\alpha+(1-t)\beta]}_{\geq 0 ??}\,\mathbf{d_1}.$$ Can you take it from here?