Find infimum of the set $A=\left \{ \frac{1}{n+1}, n \in \mathbb{N} \right \}$.

Question

Find infimum of the set $A=\left \{ \frac{1}{n+1}, n \in \mathbb{N} \right \}$.

I know that this is a basic exercise in Analysis.

I want to show that $infA=0$. For any $n\in \mathbb{N}:\frac{1}{n+1}>0,$ so $0$ is an lower bound of $A$. Let's say that there is a lower bound $L$ of $A$ such that $L>0$.

From Archimedean property there is $n_{o}\in \mathbb{N}$ such that $\frac{1}{n_{o}}<L$

So I found an element of $A$ lower than the lower bound $L$ (Proof by contradiction)

My question is: Can I use $\frac{1}{n_{o}}$ as an element of $A$ or should I write an element in the form $\frac{1}{n+1}$?

Moreover, $0$ is not min, right?

Answer 1

Since $\frac1{n+1}<\frac1n$ and $\frac1{n+1}\in A$, you're done.

And indeed $0$ is not $\min A$, since $0\notin A$.