[proofplan] We prove by contradiction. If $\mathbb{N}$ were bounded above in $F$, the least upper bound property would provide $\alpha := \sup \mathbb{N}$. Then $\alpha - 1$ is not an upper bound for $\mathbb{N}$, so some $n \in \mathbb{N}$ satisfies $n > \alpha - 1$. But then $n + 1 > \alpha$ with $n + 1 \in \mathbb{N}$, contradicting the fact that $\alpha$ is an upper bound. [/proofplan] [step:Assume $\mathbb{N}$ is bounded and derive $\alpha = \sup \mathbb{N}$] Suppose for contradiction that $\mathbb{N}$ is bounded above in $F$: there exists $M \in F$ with $n \le M$ for all $n \in \mathbb{N}$. The set $\mathbb{N}$ is nonempty ($1 \in \mathbb{N}$) and bounded above. By the least upper bound property of $F$, the supremum \begin{align*} \alpha := \sup \mathbb{N} \end{align*} exists in $F$. [guided] This is the only step that uses the least upper bound property (completeness). The Archimedean property is strictly weaker than completeness — for example, $\mathbb{Q}$ is Archimedean but does not satisfy the least upper bound property. The point of this theorem is that completeness *forces* the Archimedean property: you cannot have a complete ordered field that admits infinitesimals. [/guided] [/step] [step:Show $\alpha - 1$ is not an upper bound, producing $n > \alpha - 1$] Since $\alpha = \sup \mathbb{N}$, the value $\alpha - 1 < \alpha$ is not an upper bound for $\mathbb{N}$ (because $\alpha$ is the *least* upper bound). Therefore there exists $n \in \mathbb{N}$ with \begin{align*} n > \alpha - 1. \end{align*} [/step] [step:Reach a contradiction with $n + 1 \in \mathbb{N}$ exceeding $\alpha$] Since $n \in \mathbb{N}$, we have $n + 1 \in \mathbb{N}$ (the natural numbers are closed under successor). From $n > \alpha - 1$, adding $1$ to both sides gives \begin{align*} n + 1 > \alpha. \end{align*} But $\alpha = \sup \mathbb{N}$ is an upper bound for $\mathbb{N}$, so $n + 1 \le \alpha$ for all $n + 1 \in \mathbb{N}$. This contradicts $n + 1 > \alpha$. Therefore the assumption that $\mathbb{N}$ is bounded above is false, and $F$ is Archimedean. [guided] The proof is a clean two-line contradiction once the supremum is in hand. The mechanism: if $\alpha$ is the supremum of $\mathbb{N}$, then $\alpha - 1$ fails to be an upper bound, producing a natural number $n$ above $\alpha - 1$. But then $n + 1$ — which is also a natural number — exceeds $\alpha$ itself, violating the upper bound property of $\alpha$. The crucial fact is that $\mathbb{N}$ is closed under the successor operation $n \mapsto n + 1$, which shifts every element by exactly $1$. This rigidity means that if $\mathbb{N}$ gets within distance $1$ of its supremum, it must overshoot. [/guided] [/step]