Using the supremum axiom

Question

How can I prove that $$\sup\left\{1-\frac1n:n\in\mathbb{N}\right\}=1$$ using only the supremum definition. I've already proof that 1 is an upper bound but I don't know how to proof that this is the smallest one.

Answer 1

Take $a<1$. Then$$1-\frac1n>a\iff\frac1n<1-a\iff n>\frac1{1-a}.$$By the Archimedean property, there must be such an $n\in\mathbb N$. So, I proved that if $a<1$, then $a$ is not an upper bound. Therefore, $1$ is the least upper bound.

Answer 2

For your question : we just show that $sup\bigg\{1-{\dfrac{1}{n}}:n\in\mathbb{N}\bigg\}=1$ since you know that $1$ is an upper bound for the given set by your working.

Use the fact :For any non-empty A and B,one has \begin{align*} sup(A+B)=sup(A)+sup(B) \end{align*}

Now take $A=\bigg\{1 :n\in\mathbb{N}\bigg\}$ and $B=\bigg\{-\dfrac{1}{n}:n\in\mathbb{N}\bigg\}$.

So,it is clearly to see that $sup(A)=1$ and $sup(B)=0$.

Therefore, $sup\bigg\{1-{\dfrac{1}{n}}:n\in\mathbb{N}\bigg\}=sup(A+B)=sup(A)+sup(B)=1+0=1$

Remark : $0=inf\bigg\{\dfrac{1}{n}:n\in\mathbb{N}\bigg\}$ $\Longrightarrow$ $0=-inf\bigg\{\dfrac{1}{n}:n\in\mathbb{N}\bigg\}=sup\bigg\{-\dfrac{1}{n}:n\in\mathbb{N}\bigg\}$