The boundness of a polyhedral set implies that $y \in \mathbb{R}^n $ is equal to zero

Question

I'm solving a problem of nonlinear programming. The problem says:

Let $S_1=\{x:A_1 x\le b_1\}$ and $S_2=\{x:A_2 x\le b_2\}$ be nonempty. Define $S=S_1\cup S_2$ and $S'=\{x: x=y+z, A_1y\le b_1\lambda_1, A_2z\le b_2\lambda_2, \lambda_1+\lambda_2=1, \lambda_i\ge 0\}$. Assuming that $S_1$ and $S_2$ are bounded, show that the convex hull of $S$ is $S'$.

The solution of this exercise is in the picture. I follow all the arguments, except one: the boundness of $S_1$ implies that $y=0$.

I'm asking to you hints to understand that argument.

THANK YOU!

Answer 1

Suppose $y$ satisfies $A_1y\le b_1\lambda_1$ for $\lambda_1=0$. Then for any element $z\in S_1$, it holds that $z+cy \in S_1$ for all $c\in\mathbb{R}_+$ (because $A_1(z+cy) = A_1z + cA_1y \leq b_1 + 0 = b_1$).