Problem. Let $A = \{1 + \frac{2}{n} : n \text{ is a natural number} \}$. Find $\sup A$, with justification.
Isn't it just $3$ because $n$ is a natural number, and the lowest natural number is $1$, so the sup of this set should be $1+\frac{2}{1} = 3$?
On
You're correct that it's 3! There's a couple ways to justify this rigorously. Method (1) below is the most general way that can be used for these types of questions. Method (2) is not as general but is shorter.
Method (1): First we need to show 3 is an upper bound of $A$. This is evident by noticing $\frac 2 n \leq 2$ for all $n$, so $1+\frac 2 n \leq 1 + 2 = 3$ for all $n$, giving us the first condition. The second condition to show is that 3 is the least upper bound, not just any upper bound. Suppose for contradiction there exists a lesser upper bound $s<3$. We need to show there exists an element of $A$ greater than $s$. The element is actually $3$ occurring at $n=1$, so we have a contradiction. In the end, we have that 3 is the least upper bound of $A$, which is exactly the definition of supremum.
Method (2): It can be proven that if a set $S$ has a maximum element, then it is equal to its supremum: $\max S = \sup S$. Here, we have $\max A = 3$ since $3 \in A$ and $3 \geq 1 + \frac 2 n$ for all $n$. Thus, $\sup A = \max A = 3$.
Yep, precisely. $2/n\leq2,\forall n\in\mathbb{N}$ with equality only when $n=1$, and so $2/n+1$ has least upper bound $2+1=3$. Indeed, $s=sup(A)<3$ fails to be a least upper bound since $s<1+2/1$, and $1\in\mathbb{N}$.