I am having some troubles on a real analysis exercise where I had to prove that this set has no lower bound. Here's the problem and how I tackled the problem but I don't know what to do next to prove the statement. Any help?
Show that the set $A=\left\{\frac{5x-2}{x}\mid\ x\in\mathbb{R}_+\right\}$ has no lower bound
Proof:
BWOC let's assume that the set A has a lower bound. That means that there exists a real number m such that $m\le\frac{5x-2}{x}$ for all $x\in\mathbb{R}_+$ We can assume $m\lt 5$ because the upper bound of the set A is 5 (I checked this while proving the upper bound of this set)
(Scratch work: $m\le\frac{5x-2}{x}$
$\Leftrightarrow\ x\le\frac{5x-2}{m}=\frac{5}{m}x-\frac{2}{m}$
$\Leftrightarrow\ x-\frac{5}{m}x=x\left(1-\frac{5}{m}\right)\le-\frac{2}{m}$
$\Leftrightarrow\ x\le-\frac{2}{m}\cdot\frac{1}{1-\frac{5}{m}}=-\frac{2}{m-5}$
Here we get the idea to choose a number x as $-\frac{2}{m-5}$ Scratch work ends)
Let's choose x as $x=-\frac{2}{m-5}$
Now substituting that into $\frac{5x-2}{x}$ we get $\frac{5\left(-\frac{2}{m-5}\right)-2}{-\frac{2}{m-5}}=\frac{-\frac{10}{m-5}-2}{-\frac{2}{m-5}}=\frac{\frac{10}{m-5}+2}{\frac{2}{m-5}}=\frac{\frac{10}{m-5}+\frac{2\left(m-5\right)}{m-5}}{\frac{2}{m-5}} =\frac{\frac{10+2\left(m-5\right)}{m-5}}{\frac{2}{m-5}}=\frac{10+2\left(m-5\right)}{2}=5+m-5=m$
So $\frac{5x-2}{x}=m$
What am I supposed to do with the information that that expression is equal to m?
How do I complete my proof by contradiction?