Prove that a given set $S$ is convex

Question

Prove that the given set $$S = \left\{(x_1, x_2, x_3) \in \mathbb{R}^3 \middle| 0 \le x_1 \le x_2 \le x_3 \le 6\right\}$$ is convex.

My proof:

Let $x,y \in S$ and let $t \in [0,1]$

To prove that $S$ is convex, it suffices to show that $$tx+(1-t)y \in S$$.

However, this is where I get stuck. How can I use the constraints $0 \le x_1 \le x_2 \le x_3 \le 6$ to form my argument?

Answer 1

Let $x=(x_1, x_2, x_3)$ and $y=(y_1, y_2, y_3)$ be two points in $S,$ and let $t\in [0, 1].$ Note that $x_1, y_2\ge 0$ and hence $tx_1+(1-t)y_1\ge 0.$ Since $x_1\le x_2$ and $y_1\le y_2$ we get $$tx_1+(1-t)y_1\le tx_2+(1-t)y_2.$$ Exactly the same argument also shows that $tx_2+(1-t)y_2\le tx_3+(1-t)y_3.$ Lastly, since $x_3, y_3\le 6$ we have $tx_3+(1-t)y_3\le 6.$ This shows that $tx+(1-t)y\in S.$

Answer 2

Let $x=(x_1, x_2, x_3)$ and $y=(y_1, y_2, y_3)$. Since $t\geq 0$ and $(1-t)\geq 0$, $$t\cdot 0\le t\cdot x_1\le t\cdot x_2\le t\cdot x_3\le t\cdot 6\tag{1}$$ $$(1-t)\cdot 0\le (1-t)\cdot y_1\le (1-t)\cdot y_2\le (1-t)\cdot y_3\le (1-t)\cdot 6\tag{2}$$ By adding equation $(1)$ and $(2)$, we have $$t\cdot 0+(1-t)\cdot 0\le t\cdot x_1+(1-t)\cdot y_1\le t\cdot x_2+(1-t)\cdot y_2\le t\cdot x_3+(1-t)\cdot y_3\le t\cdot 6+(1-t)\cdot 6$$ Simplifying $$0\le t\cdot x_1+(1-t)\cdot y_1\le t\cdot x_2+(1-t)\cdot y_2\le t\cdot x_3+(1-t)\cdot y_3\le 6$$ Then $$tx + (1-t)y \in S, \quad t\in[0,1]$$ $S$ is convex.