Using Archimedian property to prove that infimum of set is $0$ Given set A = [ $\frac{1}{n} | n \in \mathbb{N}$]
It seems to me i have to prove two things :


*

*$\frac{1}{n} \geq 0 , \forall n \in \mathbb{N}$
2, Assuming $b$ be another lower bound , and so $b \leq 0$
Work :
1 Now $\frac{1}{n} \geq 0 , \forall n \in N$
Also $n \geq 1$
So $\frac{1}{n} \leq 1$ . But $\frac{1}{n} \in R^+$. So, $0 < \frac{1}{n} \leq
 1$


*Assume that $b > 0$
so by Archimedian property $\exists n \in \mathbb{N}$
 such that $\frac{1}{n} < b$ which is contradiction to the fact that b is lower bound for the set
Is this correct ?

$0$ is a lower bound, $ 1/n >0$, $n \in \mathbb{Z^{+}}$.
Assume $b >0$, real,  is a lower bound.
Archimedean principle:
There is a $n_0 \in \mathbb{Z^+}$ s.t. 
$n_0  >1/b$, i.e.
$b > 1/n_0$, a contradiction(Why?).
Hence $\inf (A)=0$. 
