Finding supremum and infimum of a set using definition of sup and inf Let $A = \{ 1 + \frac{1}{n} : n \in \mathbb{N} \} $. I want to find supremum and infimum of $A$. Intuitively, we guess $\sup A = 2 $ and $\inf A = 1 $.
First, I want to show that $2$ is upper bound of $A$. Indeed since $1/n \leq 1 $, we have that $1 + \frac{1}{n} \leq 2 $. Next, if $u$ is another upper bound of $A$, we must show that $2 \leq u $. Here is where I get stuck. How can I show this?
As for the infimum, we know $1/n > 0 $ and so $1 + \frac{1}{n} > 1 $ so $1$ is a lower bound of $A$. Suppose $l$ is another lower bound of $A$, we have to show that $1 \geq l $. How can I show this? thanks
 A: First, let's think about the definition of an upper bound of a set.  $x$ is the upper bound of a set $S$ if and only $x \geq y$ for any $y \in S$.  In this case, notice that $2 \in A$, so $x \geq 2$ for any upper bound $x$, which is the part of the proof you were missing.
The second part is a bit trickier. Suppose that there is a lower bound $\ell > 1$.  We claim that this isn't possible.  Note that $\ell - 1 > 0$, so we claim that there is some integer $N$ such that $\frac{1}{N} < \ell - 1$.  There are a lot of ways you can go about seeing this, but intuitively you can think about this is as simply saying that the sequence $\{\frac{1}{k}\}$ converges to 0.  However, this is a contradiction, as $1 + \frac{1}{N} \in A$ but $1 + \frac{1}{N} < \ell$ and  $\ell$ is supposed to be a lower bound.  Thus, for any lower bound, $\ell \leq 1$.
A: If $u$ is an upper bound of $A$ such that $u<2$, then $1+\frac{1}{n}\leq u < 2$ for all $n \in \mathbb{N}$, putting $n=1$, $2\leq u <2$ which is not true. So, $u\ge 2$.
Further, the Archimedian property may be useful if you proceed by contradiction for proof of the second. (Given a real $x$, there exists $n \in \mathbb{N}$ such that $n>x$.)
