Formal proof for supremum of $A=\{1+\frac{1}{n}:n\in\mathbb{N}\}$ is $2$.

Question

Let $A=\{1+\frac{1}{n}:n\in\mathbb{N}\}$. By intuition, I know the supremum is $2$. I want to prove that $2$ is $\sup(A)$ formally using this definition:

Let $A\subseteq \mathbb{R}$ be bounded above set. Then $u$ is called least upper bound (supremum) of $A$ if:
(i) $u$ is upper bound of $A$, i.e. $a\leq u$, for all $a\in A$.
(ii) $u$ is least upper bound of $A$, i.e. for all $\varepsilon>0$, there exists $c\in A$ such that $u-\varepsilon<c.$

For (i), we know that $\frac{1}{n}\leq 1$, for all $n\in\mathbb{N}$. So, we can get $1+\frac{1}{n}\leq 2$, for all $n\in\mathbb{N}$. From this, we have proved that $2$ is upper bound of $A$.

For (ii). Let $\varepsilon>0$. I want to show that there exists $n\in\mathbb{N}$ such that $2-\varepsilon<1+\frac{1}{n}$. This is my attempt: \begin{align*} 2-\varepsilon<1+\frac{1}{n}\\ 1-\varepsilon<\frac{1}{n} \end{align*}

I want to manipulate $1-\varepsilon<\frac{1}{n}$ such that I can use Archimedian Property, but I got stuck there.

Can anyone show that $2$ is least upper bound of $A$?

Answer 1

You do not even need the Archimedean property. One can show that if a set $A$ has a maximum $m$, then $\sup A = m$. You can easily prove this by seeing that $m$ is an upper bound and if $u$ is an upperbound for $A$ then obviously $m\le u$ because $m \in A$.

Answer 2

Firstly, $\forall n\in \Bbb N: n\geq 1\Rightarrow \frac 1n\leq 1\Rightarrow 1+\frac 1n \leq 2$, so $2$ is an upper bound of $A$. But $2\in A$, so $maxA=2$, which means that $supA=2$.

Answer 3

Your definition (ii) isn't correct - you have changed the $u$ to $a$, and also letting $a=c$ makes the definition automatically true.

A better definition is that $u$ is a least upper bound if, $\forall c$ with $c$ as an upper bound, $u\le c$.

Your sequence is readily solved by noting that it is monotonic and tightly bounded above by $2$.