[proofplan]
We prove a cycle of implications $(1) \Rightarrow (3) \Rightarrow (4) \Rightarrow (2) \Rightarrow (1)$, then show $(1) \Leftrightarrow (5)$ and $(4) \Leftrightarrow (1)$ to close the equivalence. Each implication is a short algebraic argument using only the ordered field axioms.
[/proofplan]
[step:$(1) \Rightarrow (3)$: Unboundedness of $\mathbb{N}$ implies the classical Archimedean principle]
Assume (1): for every $x \in F$, there exists $n \in \mathbb{N}$ with $n > x$. Let $x, y \in F$ with $x > 0$. We must find $n \in \mathbb{N}$ with $nx > y$.
Since $x > 0$, the element $y/x$ is well-defined in $F$. By (1), there exists $n \in \mathbb{N}$ with $n > y/x$. Multiplying both sides by $x > 0$ (which preserves the inequality in an ordered field), $nx > y$.
[guided]
The classical form (3) says "enough copies of any positive quantity exceed any target." This follows immediately from the unboundedness of $\mathbb{N}$ once we scale: asking $nx > y$ is equivalent to asking $n > y/x$ (valid since $x > 0$), and (1) guarantees such $n$ exists.
[/guided]
[/step]
[step:$(3) \Rightarrow (4)$: The classical form implies $1/n \to 0$]
Assume (3). Let $\varepsilon > 0$ in $F$. Apply (3) with $x = \varepsilon$ and $y = 1$: there exists $n \in \mathbb{N}$ with $n\varepsilon > 1$. Since $n \ge 1 > 0$, dividing both sides by $n$ gives $\varepsilon > 1/n$, i.e., $1/n < \varepsilon$.
[guided]
Statement (4) is the "epsilon" version: for any positive tolerance $\varepsilon$, the sequence $1/n$ eventually drops below $\varepsilon$. We get this from (3) by choosing $x = \varepsilon$ and $y = 1$. The resulting $n$ satisfies $n\varepsilon > 1$, so $1/n < \varepsilon$. Note that $n \in \mathbb{N}$ ensures $n \ge 1 > 0$, so division by $n$ is valid and preserves the inequality direction.
[/guided]
[/step]
[step:$(4) \Rightarrow (2)$: $1/n \to 0$ implies no positive infinitesimals]
Assume (4). Suppose for contradiction that there exists $x \in F$ with $0 < x < 1/n$ for all $n \in \mathbb{N}$. Set $\varepsilon := x > 0$. By (4), there exists $n \in \mathbb{N}$ with $1/n < \varepsilon = x$. But we assumed $x < 1/n$ for all $n \in \mathbb{N}$, giving $x < 1/n < x$, a contradiction. Therefore no such $x$ exists.
[/step]
[step:$(2) \Rightarrow (1)$: Absence of infinitesimals implies $\mathbb{N}$ is unbounded]
Assume (2). Suppose for contradiction that $\mathbb{N}$ is bounded above in $F$: there exists $M \in F$ with $n \le M$ for all $n \in \mathbb{N}$. Then for all $n \in \mathbb{N}$, $1/(M+1) < 1/n$ is not necessarily immediate; instead we argue as follows.
Since $n \le M$ for all $n \in \mathbb{N}$, we have $n + 1 \le M + 1$ for all $n$, so $n \le M$ for all $n$. In particular, $M \ge 1 > 0$. For any $n \in \mathbb{N}$, $n \le M$ gives $1/M \le 1/n$ (since $n, M > 0$ and $n \le M$ implies $1/n \ge 1/M$). Therefore $0 < 1/M \le 1/n$ for all $n \in \mathbb{N}$, and $1/(M+1) < 1/M \le 1/n$ for all $n$. Set $x := 1/(M+1) > 0$. Then $x < 1/n$ for all $n \in \mathbb{N}$, so $x$ is a positive infinitesimal, contradicting (2).
[guided]
We prove the contrapositive: if $\mathbb{N}$ is bounded, then a positive infinitesimal exists. If $n \le M$ for all $n \in \mathbb{N}$, then $M \ge 1 > 0$ (since $1 \in \mathbb{N}$). For every $n \in \mathbb{N}$, $n \le M$ implies $1/n \ge 1/M$ (both sides are positive, and taking reciprocals reverses the inequality). Define $x := 1/(M+1)$. Since $M + 1 > M \ge n$ for all $n$, we get $x = 1/(M+1) < 1/n$ for all $n \in \mathbb{N}$. Also $x > 0$ since $M + 1 > 0$. So $x$ is a positive element smaller than every $1/n$, violating (2).
[/guided]
[/step]
[step:$(1) \Rightarrow (5)$: Unboundedness of $\mathbb{N}$ implies $\mathbb{Z}$ is cofinal in both directions]
Assume (1). Let $x \in F$. By (1), there exists $n \in \mathbb{N}$ with $n > x$. Set $M := n \in \mathbb{Z}$; then $M > x$.
For the lower bound, apply (1) to $-x$: there exists $m' \in \mathbb{N}$ with $m' > -x$, i.e., $-m' < x$. Set $m := -m' \in \mathbb{Z}$; then $m < x$.
Therefore $m < x < M$ with $m, M \in \mathbb{Z}$.
[/step]
[step:$(5) \Rightarrow (1)$: Cofinality of $\mathbb{Z}$ implies unboundedness of $\mathbb{N}$]
Assume (5). Let $x \in F$. By (5), there exists $M \in \mathbb{Z}$ with $M > x$. If $M \ge 1$, then $M \in \mathbb{N}$ (since $\mathbb{N} = \{1, 2, 3, \ldots\} \subset \mathbb{Z}$ consists of the positive integers), and $n := M$ satisfies $n > x$. If $M \le 0$, then $n := 1 \in \mathbb{N}$ satisfies $n = 1 > 0 \ge M > x$ — wait, we only know $M > x$, not $M > 0$. Let us re-examine: $M > x$, so if $M \le 0$, then $x < M \le 0 < 1$, so $n = 1 > x$. In either case, there exists $n \in \mathbb{N}$ with $n > x$.
[/step]
