Every argument in real analysis that begins "let $\varepsilon > 0$" depends, at its core, on being able to find a natural number large enough to make some quantity small. If we want $1/n < \varepsilon$, we need an $n \in \mathbb{N}$ with $n > 1/\varepsilon$. This is not a logical triviality — it is a statement about the *structure* of $\mathbb{R}$ and its relationship to $\mathbb{N}$. There exist ordered fields in which no such $n$ exists: fields containing "infinitely large" elements that exceed every natural number, or equivalently, "infinitesimally small" positive elements that are less than $1/n$ for every $n$. The **Archimedean property** is the axiom that rules out this pathology. It asserts that the natural numbers are cofinal in $\mathbb{R}$: no real number is an upper bound for $\mathbb{N}$.

What makes this property remarkable is that it is *not* an independent axiom — it follows from [completeness](/page/Supremum%20and%20Infimum). Any ordered field that satisfies the least upper bound property is automatically Archimedean. This means that the Archimedean property is, in a precise sense, a *consequence* of the completeness of $\mathbb{R}$, and the fact that so much of analysis rests on it is really a reflection of how deeply completeness pervades the subject.

[example: Failure In Hyperreals] The field of hyperreal numbers ${}^{*}\mathbb{R}$ from nonstandard analysis is an ordered field extension of $\mathbb{R}$. It contains an element $\omega$ that is *infinite*: $\omega > n$ for every $n \in \mathbb{N}$. Consequently, the element $\varepsilon = 1/\omega$ satisfies $0 < \varepsilon < 1/n$ for every $n \in \mathbb{N}$ — it is a positive infinitesimal. The hyperreals are a non-Archimedean ordered field. They are not complete: the set $\mathbb{N} \subset {}^{*}\mathbb{R}$ is bounded above (by $\omega$) but has no least upper bound in ${}^{*}\mathbb{R}$. [/example]

This example shows that the Archimedean property genuinely constrains the structure of an ordered field. Without it, the relationship between "large" and "small" breaks down: there exist positive quantities that no finite sum of copies can bring above $1$, and quantities that exceed every finite threshold.

## Definition

The Archimedean property can be stated in several equivalent forms. The first is the most common; the others are frequently more useful in practice.

[definition: Archimedean Property] An ordered field $(F, +, \cdot, <)$ is **Archimedean** if for every $x \in F$, there exists $n \in \mathbb{N}$ such that $n > x$. Equivalently, $\mathbb{N}$ is unbounded in $F$: no element of $F$ is an upper bound for the set $\{1, 2, 3, \ldots\} \subset F$ (where we identify $n \in \mathbb{N}$ with the $n$-fold sum $1_F + 1_F + \cdots + 1_F$). [/definition]

The identification of $n \in \mathbb{N}$ with an element of $F$ uses the unique ordered-field homomorphism $\mathbb{Z} \to F$ mapping $1 \mapsto 1_F$, extended to $\mathbb{Q} \hookrightarrow F$ by $p/q \mapsto p \cdot q^{-1}$. In an Archimedean ordered field, this embedding is always injective (since $n > 0$ in $F$ for all $n \in \mathbb{N}$), so $\mathbb{Q}$ embeds as an ordered subfield of $F$.

The following reformulations are all equivalent to the definition above and appear frequently in practice.

[quotetheorem:1232]

The equivalences are straightforward: (1) and (4) are contrapositives of each other (set $\varepsilon = 1/x$ when $x > 0$), (1) and (3) are equivalent by dividing by $x$, and (2) is the negation of the failure of (4). Form (3) is closest to the historical statement attributed to Archimedes: given any two lengths, sufficiently many copies of the smaller exceed the larger. Form (4) is the version that appears implicitly in every $\varepsilon$-$\delta$ argument.

The critical point is that Form (2) shows the Archimedean property is equivalent to the absence of infinitesimals. An ordered field is non-Archimedean if and only if it contains a positive element smaller than every $1/n$ — an infinitesimal. This is the fundamental dichotomy: either the field has infinitesimals, or $\mathbb{N}$ is unbounded.

## The Archimedean Property From Completeness

The most important question about the Archimedean property is: *where does it come from?* Can we derive it, or must we assume it as an axiom?

In the standard construction of $\mathbb{R}$ — whether by Dedekind cuts or Cauchy completion of $\mathbb{Q}$ — the [least upper bound property](/page/Supremum%20and%20Infimum) (completeness) is the defining characteristic of $\mathbb{R}$. The Archimedean property turns out to be a *consequence* of completeness, not an independent assumption. This is a nontrivial fact: completeness is a statement about the existence of suprema, while the Archimedean property is a statement about the growth of $\mathbb{N}$. That the former implies the latter reveals a deep structural constraint.

[quotetheorem:1233]

The argument is a proof by contradiction that illustrates the power of the completeness axiom. If $\mathbb{N}$ were bounded above in $F$, then by completeness, $\alpha = \sup \mathbb{N}$ would exist. But then $\alpha - 1$ would not be an upper bound for $\mathbb{N}$ (since $\alpha$ is the *least* upper bound), so there would exist $m \in \mathbb{N}$ with $m > \alpha - 1$, giving $m + 1 > \alpha$. Since $m + 1 \in \mathbb{N}$, this contradicts $\alpha$ being an upper bound for $\mathbb{N}$.

This argument deserves reflection. The key step is that $\mathbb{N}$ is *closed under successor*: if $m \in \mathbb{N}$, then $m + 1 \in \mathbb{N}$. This is what prevents $\mathbb{N}$ from having a supremum in $F$ — any candidate supremum $\alpha$ can be "exceeded" by stepping from some $m > \alpha - 1$ to $m + 1 > \alpha$. Completeness thus forces $\mathbb{N}$ to be unbounded, because the successor operation provides an inexhaustible supply of elements that push past any proposed bound.

[remark: The Converse Fails] The Archimedean property does *not* imply completeness. The ordered field $\mathbb{Q}$ is Archimedean — for any $x \in \mathbb{Q}$, the ceiling $\lceil x \rceil + 1$ is a natural number exceeding $x$ — but $\mathbb{Q}$ does not satisfy the least upper bound property: the set $\{q \in \mathbb{Q} : q^2 < 2\}$ is bounded above but has no supremum in $\mathbb{Q}$. Thus the Archimedean property is strictly weaker than completeness. [/remark]

This asymmetry is important for the foundations of analysis. Completeness is the axiom that distinguishes $\mathbb{R}$ from $\mathbb{Q}$; the Archimedean property is a consequence that $\mathbb{R}$ shares with $\mathbb{Q}$ but not with fields like the hyperreals. The logical relationship is: