Use the Archimedean Property of R and the definition of the infinum to prove...

Question

that $\inf\left\{\frac2n\,\middle|\,n ∈ \Bbb N\right\}= 0.$ (where $\Bbb{N}$ is the set of all positive integers)

Here is what I have so far:

$ 0 < 2/n\Rightarrow 0\;$ is a lower bound.

Let $a > 0 \Rightarrow$ by the Archimedean Property $\;\exists n \in N:1/n < a$

For the next step I believe I have to conclude that $a$ is not a lower bound, and then conclude that $0$ is the greatest lower bound, but I am not sure how to do so.

Answer 1

Since you are dealing with the numbers of the form $\frac2n$, you should apply the Archimedean Property to $\frac a2$. So, there's some $n\in\mathbb N$ such that $\frac1n<\frac a2$, which is equivalent to $\frac2n<a$. So, $a$ is not a lower bound of your set and therefore the greatest lower bound (that is, the infimum) is $0$.

Answer 2

To show that the infimum is $0$, we first observe that $\frac2n$ can't be negative for any values of $n$, so the infimum likewise must be greater than or equal to $0$.

Next, we assume for the sake of contradiction that the infimum $=\epsilon>0$. Then, by the archimedian property, we know that $\exists n\in\mathbb N$ such that $n>\frac1\epsilon$. This implies that $\frac1n<\epsilon$, which is a contradiction. Hence we have completed the proof.

Answer 3

Show that $A:= \inf ${$2/n| n \in \mathbb{Z^+}$} $= 0$.

1) $0$ is a lower bound , since for

a $2/n \in A$ , we have $0 <2/n$.

2) $0$ is greatest lower bound.

Assume there is a lower bound $b >0$.

Archimedean principle:

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_0 > 2/b$.

For $n \ge n_0 :$

$b>2/n_0 \ge 2/n >0$.

$b>0$ is not a lower bound.

Hence $\inf (A) =0$.