I'm given these two sets
$A\subset (0,+\infty ),$ inf$A=0$ and $A$ is not upper bounded
$B=\left \{ \frac{x}{x+1}:x\in A \right \}$
and I have to find the supremum.
Here's the solution my book gives to prove that $supB=1$:
If $y\in B$ then there is a $x \in A$ s.t. $y=\frac{x}{x+1}<1$
We choose $ε>0$ and we have to find $x\in A$ such that $\frac{x}{x+1}<1-ε$ that is $x>\frac{1}{ε}-1$ ...
the proof continues but my problem is this segment, how does it go from here $\frac{x}{x+1}<1-ε$ to here $x>\frac{1}{ε}-1$ ?
Note that $$ \frac{x}{x+1} = \frac{x+1-1}{x+1} = 1- \frac{1}{x+1}, $$ so $$ \frac{x}{x+1} < 1 - \epsilon \quad \Leftrightarrow \quad 1-\frac{1}{x+1} < 1 - \epsilon \quad \Leftrightarrow \quad \frac{1}{x+1} > \epsilon, $$ and inverting this gives (this is allowed since $x > 0,$ so $x+1 > 1 > 0)$ $$ x+1 < \frac{1}{\epsilon} \quad \Leftrightarrow \quad x < \frac{1}{\epsilon}-1. $$