Help to prove the condition that a right half-open interval is not empty The right half-open interval is defined as:
$[a,b) = \{x \in \mathbb{R}|a \le x \lt b\}$
I need to prove:
$[a,b) \ne \emptyset \iff a<b$
My attempt:
For $\Rightarrow$:
$$\begin{align}
[a,b)\ne \emptyset & \Rightarrow & a\le x \lt b \\
&\Rightarrow & a\le x \text{ and } x \lt b \\
& \Rightarrow & a \lt b
\end{align}$$
For $\Leftarrow$:
$\begin{align}
a<b & \Rightarrow & 2a < a+b \text{ and } a+b < 2b\\
& \Rightarrow & 2a<a+b<2b \\
& \Rightarrow & a< \frac{a+b}{2} < b \\
& \Rightarrow & (a,b)
\end{align}$
I don't know how to start with $a<b$ and imply $a\le x <b$. I face similar problem too with a left half-open interval, $(a,b]$
 A: Starting from your definition of a right-open interval
$$ [a\,,\,b] = \{x \in \mathbb{R} \,|\, a \leq x < b \} $$
We want to prove
$$ [a , b] \neq \emptyset \Leftrightarrow a < b $$
$\textbf{Case } (\Rightarrow) $ :
As $[a\,,\,b) \neq \emptyset$, there exists $x \in \mathbb{R}$ such that $x \in [a\,,\,b)$. This $x$ satisfies the condition
$$ a \leq x < b $$
Substracting by $a$ on both sides, we obtain
$$ 0 \leq x - a < b -a \;\; \Rightarrow 0 < b - a$$
so that we have $a < b$. ($\Rightarrow$) is proved.
$\textbf{Case } (\Leftarrow)$ :
We consider $x = \frac{a+b}{2} $. Let's show that we have $a \leq x < b $.


*

*1/ We start from the inequality $a < b$. Adding $a$ on both sides and dividing by $2$, we obtain
$$ a < x $$
This condition straightforwardly implies that $a \leq x$

*2/ Starting from the inequality $a < b$, we add $b$ on both sides, and divide by $2$, so that we obtain
$$ x < b $$
The two results (1) and (2) therefore leads us to have
$$ a \leq x < b $$
As a consequence, we have $x \in [a\,,\,b)$. ($\Leftarrow$) is proved. 
