Prove $\{x \in \mathbb{R}:a\leq x\leq b\}=\{y\in \mathbb{R}:\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb\}$ I'm doing the $\supseteq$ direction first as I find that the easiest.
Let $A={\{x \in \mathbb{R}:a\leq x\leq b}\}$
Let $B=\{y\in \mathbb{R}:\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb\}$
Suppose y is an arbitrary element of B. Then $\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb$
then $sa+ta=a$ and $sb+tb=b$. 
Since $a\leq b$, $tb\geq ta$ then $sa+tb\geq a$
And $sa\leq sb$ so $a\leq sa+tb \leq b$
Since $y=sa+tb$ then $y \in [a,b]$ as required.
For the $\subseteq$ direction I feel I didn't do this right.
Let A,B as above. Suppose x is an arbitrary element of A. Then $a\leq x\leq b$.
Suppose $\exists s,t\in [0,1]\; with\; s+t=1$.
then $a=sa+ta \; and \; sb+tb=b$
Since $a\leq b$, $sa\leq sb \; and \; ta\leq tb$
then $a=sa+ta\leq sa+tb\leq sb+tb=b$ 
then $sa+tb$ is an arbitrary element of A. Thus $x=sa+tb$ and $x\in B$.

One direction is straightforward.
We have $a = sa+ta \le sa+tb \le sb+tb = b$.
 Hence $sa+tb $ is in the first set (A).
The other direction just needs a little trick, suppose $x$ is in the first set, let
$s = {x-a \over b-a}$, $t = {b-x \over b-a}$. Then $s,t$ are non negative, $s+t = 1$ and $x = sa+tb$. Hence $x$ is in the second set.
