$A = \left\{ \frac{x}{x+1} : x > 0 \right\}$. Prove inf $A$ = $0$.

$A = \left\{ \frac{x}{x+1} : x > 0 \right\}$

The set A is bounded, because $0 < 1 - \frac{1}{x+1} < 1$ for all $ x > 0 $.

My attempt:

inf $A$ = $0$, because:

1. $ 0 < 1 - \frac{1}{x+1}$ for all $ x > 0 $, so $ 0 $ is a lower bound of $A$.

2. For any $ w > 0 $, $w$ is not a lower bound of $ A $. By the Archimedean property, there exists $ x_0 \in \mathbb{N} $ such that $\frac{1}{x_0} < w $. Consequently, there exists $a_1 = 1 - \frac{1}{x_0 + 1}\in A$ such that:

   $ a_1 = 1 - \frac{1}{x_0+1}<\frac{1}{x_0} < w$.

Is this step correct? In no. 2 want to show there exist $a_1∈A$ such that $a_1<w$.