Using Archimedian property to prove that infimum of set is $0$

Question

Given set A = [ $\frac{1}{n} | n \in \mathbb{N}$]

It seems to me i have to prove two things :

1. $\frac{1}{n} \geq 0 , \forall n \in \mathbb{N}$

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, $0 < \frac{1}{n} \leq 1$

2. 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 ?

Answer 1

$0$ 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 $\inf (A)=0$.

Answer 2

Observe that 0 is a lower bound of $\frac{1}{n}$ Archimedian property- if $x > 0 $ then there exists $n \epsilon \mathbb{R} $ such that $nx > y; x,y \epsilon \mathbb{R}$. Set $x = \epsilon$ and $y=1$ $$ \epsilon > \frac{1}{n} \forall \epsilon > 0$$ . Hence infimum of $\frac{1}{n} $ is $0$ .