Proof verification - Infimum

Question

I want to prove that $\inf A=0$ where $A=\left\{ {\dfrac{1}{n+1}} \mid n \in \mathbb{N}\right\} $.

I found an easier way to do it, but I wanted to try the $\epsilon$ Lemma along with the Archimedian Axiom.

First the Lemma: $s = \inf A \iff \forall \epsilon >0, \exists \ a\in A$ such that $\epsilon + s > a$.

I tried the following:

Let $\epsilon > 0$ be arbitrarily chosen Then choose $a = \frac{1}{y+1}$ with $y > \frac{1}{\epsilon} - 1$ and $y \in \mathbb{N}$. Such an $y$ exists because of the Archimedian property. It follows that:

$$y > \frac{1}{\epsilon} - 1 \iff y+1 > \frac{1}{\epsilon} \iff \frac{1}{1+y}< \epsilon$$

Now it follows that our infimum is $0$.

Is this correct? To argue that $0$ is a lower bound is easy, so left that out here.

Answer 1

Yes, your proof is correct. And I can't think of a significant improvement with this proof strategy, either.

Answer 2

An option:

1) $0$ is a lower bound , since $0 \lt 1/(n+1)$, $ n \in \mathbb{N}.$

Assume $a \gt 0$, $a$ real, is a lower bound:

Archimedes:

There is a $n_0 \in \mathbb{N}$ such that

$n_0 \gt 1/a.$

For $n \ge n_0$ we have

$0 \lt \dfrac{1}{n+1} \le \dfrac{1}{1+n_0} \lt a$.

Contradiction.

Hence $\inf$ {$\dfrac{1}{n+1}$| $n \in \mathbb{N}$}$=0$.