$\sup\big(\{\frac{1}{n}: n \in \mathbf{N}\} \big) = 1.$

For the definitions of an upper bound and the least upper bound of a set, see this.
For the Archimedean Principle, see this

Proof:

Let $A = \{\frac{1}{n}: n \in \mathbf{N}\}$. To prove that $\sup(A)=1$, we will show that 1 is an upper bound on $A$ and that 1 is the least upper bound on $A$ using lemma 1.3.8. First, to prove that $1$ is an upper bound, note that for all $n \in \mathbf{N}$, $n \geq 1$. Divide both sides by $n$ to get $1 \geq \frac{1}{n}$. Therefore, 1 is an upper bound on $A$.

Next, to show that $0$ is the least upper bound, we need to find an element $x \in A$ such that for any $\epsilon > 0$,

$$ \begin{align*} x > 1 - \epsilon. \end{align*} $$

This means that we’ve found an element in $A$ that results in $1$ no longer being an upper bound. Choose $x = 1 \in \mathbf{N}$ to be the element. We know that $1 \geq 1 - \epsilon$ for any $\epsilon > 0$. Therefore, $1$ is the least upper bound of $A$. $\blacksquare$

References:

Real Analysis by Jay Cummings