How do I prove that $[0,1]$ is convex?

Question

I am having trouble with bounding the upper bound. Since the requirement is

$$\forall t \in [0,1], \forall u,v \in C = [0,1] \to tu + (1-t)v \in C$$

Therefore, I need $$tu + (1-t)v \leq 1$$

but I can only bound it with $2$, as in $tu \leq 1$ and $(1-t)v \leq 1$ then $tu + (1-t)v \leq 1+ 1 = 2$.

Answer 1

Without loss of generality, let $0\le u\le v\le 1$. Then: $$\color{red}0\le u=tu+(1-t)u\le \color{red}{tu+(1-t)v}\le tv+(1-t)v=v\le \color{red}1.$$

Answer 2

Hint: use that $u, v \in C$, i.e. $\leq 1$.

Answer 3

$$tu+(1-t)v\le \max{\{tu+(1-t)u, tv+(1-t)v\}}=\max{\{u,v\}}\le 1$$ Similarly $tu+(1-t)v\ge \min{\{u,v\}}\ge 0$.

Answer 4

You have to prove that

$$tu + (1-t)v\leq 1$$

You can do that by using the facts that:

• $u\leq 1$
• $t\geq 0$
• Therefore, $ut \leq t$

and doing something similar with $(1-t)v$.

Answer 5

for showing that $C=[0.1]$ is a convex set,suppose that $0\lt t\lt1$ and $x ,y \in C$,we want to show that $tx+(1-t)y \in C$. We have$$0\leq tx+(1-t)y\leq t+(1-t)=1$$. Therefore $tx+(1-t)y\in C$