How to prove that the following is a convex set: $[3,5]\times [1,2]\times \{1\} \subseteq \mathbb{R}^3 $

Question

This might be very easy, but I can not see how to prove that:

$[3,5]\times [1,2]\times \{1\} \subseteq \mathbb{R}^3 $ is a convex set.

Thank you for your help.

Answer 1

For any $p_1,q_1\in[3,5]$ and any $t\in[0,1]$, it holds $tp_1+(1-t)q_1\in[3,5]$

This holds, because $3\le p_1,q_1 \le 5$, therefore $3= t\cdot 3+(1-t)\cdot 3\le tp_1+(1-t)q_1\le t\cdot 5+(1-t)\cdot 5=5$, this means $tp_1+(1-t)q_1\in[3,5]$

The same arguments applies for the second and third component:

For any $p_2,q_2\in[1,2]$ and any $t\in[0,1]$, it holds $tp_2+(1-t)q_2\in[1,2]$

For any $p_3,q_3\in\{1\}$ and any $t\in[0,1]$, it holds $tp_3+(1-t)q_3\in\{1\}$

Therefore:

For any $p=(p_1,p_2,p_3),q=(q_1,q_2,q_3)\in[3,5]\times[1,2]\times{1}$ and any $t\in[0,1]$, it holds $tp+(1-t)q\in[3,5]\times[1,2]\times\{1\}$