How to prove the open interval $(1,5)$ is a convex set? I want to prove the interval $(1,5)$ is a convex set. 
A convex set is a set having all the convex linear combinations of its point in it, where a convex linear combination is a linear combination of the form $X=(1-\alpha)X_1+ \alpha X_2$ where $\alpha$ lies between $0$ and $1$ and the sum of coefficients of above linear combination is equal to 1.
How can we prove the interval $(1,5)$ a convex set?
 A: You just have to prove that for every $\alpha \in (0,1)$ and every pair $x_1$, $x_2$ $\in (1,5)$, $\alpha x_1 + (1-\alpha)x_2$ also lies in $(1,5)$.
Suppose without loss of generality that $x_1 \leq x_2$. Then for every $\alpha \in (0,1)$ (which implies that both $\alpha$ and $1-\alpha$ are positive)
$\alpha x_1 + (1-\alpha)x_2 \leq \alpha x_2 + (1-\alpha)x_2 = x_2$ 
and
$\alpha x_1 + (1-\alpha)x_2 \geq \alpha x_1 + (1-\alpha)x_1 = x_1$.
Hence we find that for every $\alpha \in (0,1)$, $\alpha x_1 + (1-\alpha)x_2 \in [x_1,x_2]$.
Hence for every pair $x_1$, $x_2$, any convex-combination of $x_1$, $x_2$ lies in $[x_1,x_2]$. And since $x_1$, $x_2$ $\in (1,5)$, every convex-combination of any two points in $(1,5)$ also lies in $(1,5)$.
A: If $x_1\leq x_2$ then $$x=(1-\alpha)x_1+\alpha x_2=x_1+\alpha(x_2-x_1)$$ If $\alpha \in [0,1]$ then $$0\leq \alpha(x_2-x_1)\leq x_2-x_1$$ so: $$x_1\leq x\leq x_1+(x_2-x_1)=x_2$$
Then $x_1,x_2\in (1,5)$ gives: $$1< x_1\leq x\leq x_2<5$$
A: Hint : 


*

*$1<X_1<5\Rightarrow 1-\alpha<(1-\alpha)X_1<5(1-\alpha)$

*$1<X_1<5\Rightarrow \alpha<\alpha X_2<5\alpha$


Can you now bound $(1-\alpha)X_1+ \alpha X_2$ just by adding above two inequalities?
