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: \begin{align*} \text{Completeness} \implies \text{Archimedean property} \implies \text{No infinitesimals}, \end{align*} but neither implication reverses. ## Density of the Rationals One of the most important consequences of the Archimedean property is that $\mathbb{Q}$ is *dense* in $\mathbb{R}$: between any two distinct real numbers lies a rational number. This fact is used constantly throughout analysis — in approximation arguments, in the construction of countable dense subsets, in the definition of the Borel $\sigma$-algebra — yet its proof rests squarely on the Archimedean property, and on nothing else. The difficulty is this: given real numbers $a < b$, we need to find integers $p$ and $q > 0$ such that $a < p/q < b$. The denominator $q$ must be large enough that the "grid" of rationals with denominator $q$ — the points $\ldots, -1/q, 0, 1/q, 2/q, \ldots$ — is fine enough to place a grid point in the interval $(a, b)$. The spacing of this grid is $1/q$, so we need $1/q < b - a$, i.e., $q > 1/(b-a)$. The Archimedean property guarantees that such a $q$ exists. [quotetheorem:1234] The construction is explicit. By the Archimedean property, choose $q \in \mathbb{N}$ with $q > 1/(b-a)$, so that $1/q < b - a$. Again by the Archimedean property, the set $\{m \in \mathbb{Z} : m > qa\}$ is nonempty (since $\mathbb{Z}$ is cofinal in both directions). Let $p$ be the least such integer: $p - 1 \leq qa < p$. Then $a < p/q$, and $p/q = (p-1)/q + 1/q \leq a + 1/q < a + (b - a) = b$. Thus $a < p/q < b$, and $r = p/q \in \mathbb{Q}$ is the desired rational. Notice that this argument uses the Archimedean property *twice*: once to find $q$ (making the grid fine enough) and once to find $p$ (locating the correct grid point). It also uses the *well-ordering principle* for $\mathbb{Z}$ (to pick the least integer exceeding $qa$). The well-ordering step is essential — without it, we could find *some* integer $m > qa$, but we could not guarantee $m/q < b$. [example: Density Of Irrationals] The irrationals $\mathbb{R} \setminus \mathbb{Q}$ are also dense in $\mathbb{R}$. Given $a < b$, apply the density of the rationals to the interval $(a/\sqrt{2}, b/\sqrt{2})$ to find $r \in \mathbb{Q}$ with $a/\sqrt{2} < r < b/\sqrt{2}$. Then $a < r\sqrt{2} < b$, and $r\sqrt{2}$ is irrational (if $r \neq 0$; if $r = 0$, use $a < 0 < b$ and pick any irrational in $(a,b)$ directly). The case $r = 0$ requires $a < 0 < b$, but then the irrationals in $(a,0)$ or $(0,b)$ are already available by reapplying the argument to the appropriate subinterval. [/example] The density of both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ illustrates a fundamental feature of the real line: the rationals and irrationals are interleaved everywhere, despite having vastly different cardinalities ($\mathbb{Q}$ is countable while $\mathbb{R} \setminus \mathbb{Q}$ is uncountable). Both density results depend on the Archimedean property, and both fail in non-Archimedean fields. ## The Floor Function and Integer Parts The Archimedean property also guarantees the existence of the floor (integer part) function, a tool used so frequently that its dependence on the Archimedean property is often overlooked. The problem is this: given $x \in \mathbb{R}$, we want to find the largest integer that does not exceed $x$. This requires two things: (1) there must exist *some* integer below $x$ (guaranteed by the Archimedean property applied to $-x$), and (2) among all integers below $x$, there must be a greatest one (guaranteed by well-ordering of $\mathbb{Z}$ above any lower bound, which itself follows from the Archimedean property). [quotetheorem:1235] The integer $n$ is called the **floor** of $x$, written $\lfloor x \rfloor$. The fractional part is $\{x\} := x - \lfloor x \rfloor \in [0, 1)$. The floor function is indispensable in number theory (the study of $\lfloor x/p \rfloor$ in Legendre's formula), in analysis (truncation arguments, partitioning intervals), and in the proof of the density theorem above (where we implicitly used $p = \lfloor qa \rfloor + 1$). ## The Archimedean Property in Epsilon-Delta Arguments The pervasive role of the Archimedean property in analysis becomes visible once one examines the logical structure of convergence arguments. Every proof that a [sequence converges](/page/Convergence%20(Real%20Sequences)) or that a function is continuous at a point relies, at some step, on finding a natural number large enough to control an error term. The Archimedean property is what makes this step valid. Consider a concrete instance. To prove that $a_n = 1/n \to 0$ as $n \to \infty$, one must show: for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $|1/n - 0| < \varepsilon$ for all $n \geq N$. The critical step is choosing $N$: we need $N > 1/\varepsilon$, and the Archimedean property guarantees such an $N$ exists. Without the Archimedean property, this argument would fail — and indeed, in the hyperreals, the sequence $1/n$ (indexed over $\mathbb{N}$) does *not* converge to $0$ in the order topology, because the infinitesimal $1/\omega$ is a "limit point" that the standard natural numbers never reach. [example: Nested Interval Argument] The nested interval property — that a decreasing sequence of closed bounded intervals $[a_n, b_n]$ with $b_n - a_n \to 0$ has $\bigcap_{n=1}^\infty [a_n, b_n] = \{x\}$ for a unique $x \in \mathbb{R}$ — uses the Archimedean property in the step "$b_n - a_n \to 0$." To say the lengths shrink to zero means: for every $\varepsilon > 0$, eventually $b_n - a_n < \varepsilon$. Verifying this for a specific nested interval scheme (say, $[a_n, b_n]$ with $b_n - a_n = (b_1 - a_1)/2^n$) requires checking $(b_1 - a_1)/2^n < \varepsilon$, i.e., $2^n > (b_1 - a_1)/\varepsilon$. Since $2^n \geq n$ for all $n \geq 1$ (by induction), it suffices to find $n > (b_1 - a_1)/\varepsilon$, which is possible by the Archimedean property. [/example] More broadly, the following pattern recurs throughout analysis: 1. **Goal:** Show some error $E_n$ satisfies $E_n < \varepsilon$ for large $n$. 2. **Estimate:** Establish $E_n \leq C/n^\alpha$ for some $C > 0$, $\alpha > 0$. 3. **Archimedean step:** Choose $N \in \mathbb{N}$ with $N > (C/\varepsilon)^{1/\alpha}$. Step 3 is always justified by the Archimedean property. This pattern is so fundamental that it is rarely named; the Archimedean property operates as invisible infrastructure beneath every convergence argument. ## Connections to the Cauchy Criterion The interaction between the Archimedean property and the [Cauchy criterion](/page/Cauchy%20Sequence) deserves attention. A sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ is Cauchy if for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $|a_m - a_n| < \varepsilon$ for all $m, n \geq N$. The completeness of $\mathbb{R}$ guarantees that every Cauchy sequence converges. The Archimedean property plays two distinct roles here: **First**, it ensures that the Cauchy condition is nontrivial. In a non-Archimedean field, a sequence could have all differences $|a_m - a_n|$ bounded by an infinitesimal for $m, n$ large, yet never converge to a standard element. The Archimedean property ensures that "for every $\varepsilon > 0$" ranges over all positive reals, with no infinitesimal gap between $0$ and the positive elements. **Second**, it powers the standard technique for proving a sequence is Cauchy. The typical argument establishes $|a_m - a_n| \leq C/(m \wedge n)^\alpha$ and then uses the Archimedean property to find $N$ with $C/N^\alpha < \varepsilon$. [example: Cauchy Implies Bounded] The proof that every Cauchy sequence is bounded uses the Archimedean property. Given a Cauchy sequence $(a_n)$, take $\varepsilon = 1$. Then there exists $N$ such that $|a_m - a_n| < 1$ for all $m, n \geq N$. In particular, $|a_n| < |a_N| + 1$ for all $n \geq N$. Set $M = \max(|a_1|, |a_2|, \ldots, |a_{N-1}|, |a_N| + 1)$; then $|a_n| \leq M$ for all $n$. The step "take $\varepsilon = 1$" is innocuous in $\mathbb{R}$ because $1 > 0$, but the step "find $N$ such that..." is where the Archimedean property enters: the existence of such $N$ is part of the Cauchy hypothesis, which itself is only meaningful because the quantifier "for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$" has content in an Archimedean field. [/example] ## Non-Archimedean Ordered Fields To appreciate what the Archimedean property gives us, it is instructive to examine ordered fields where it fails. The failure always manifests in the same way: the field contains elements that are "infinitely large" (exceeding every natural number) and, reciprocally, positive elements that are "infinitesimally small" (less than $1/n$ for every $n$). ### The Field of Rational Functions The simplest example of a non-Archimedean ordered field requires no sophisticated construction. [example: Rational Functions As A Non-Archimedean Field] Let $F = \mathbb{R}(t)$ be the field of rational functions in one variable with real coefficients. Define an ordering by: $f > 0$ if $f(x) > 0$ for all sufficiently large $x \in \mathbb{R}$. Under this ordering, $t > n$ for every $n \in \mathbb{N}$, because $t - n > 0$ for all $x > n$. Thus $t$ is an "infinite" element: it exceeds every natural number. Correspondingly, $1/t$ is an infinitesimal: $0 < 1/t < 1/n$ for every $n \in \mathbb{N}$. This field is not complete: the set $S = \{n \in \mathbb{N}\} \subset F$ is bounded above (by $t$) but has no least upper bound. If $\alpha = p(t)/q(t)$ were $\sup S$, then $\alpha - 1$ would also be an upper bound for $S$ (since $\alpha - 1 > n$ for all $n$ implies $\alpha > n+1$ for all $n$), contradicting the leastness of $\alpha$. This argument mirrors the proof that completeness implies the Archimedean property, run in reverse: the non-Archimedean field necessarily lacks completeness. [/example] ### The $p$-Adic Numbers A more profound example of non-Archimedean structure arises from a fundamentally different notion of "size." [explanation: The p-Adic Absolute Value] Fix a prime $p$. Every nonzero rational number $x \in \mathbb{Q}$ can be written uniquely as $x = p^k \cdot a/b$ where $k \in \mathbb{Z}$ and $a, b$ are integers not divisible by $p$. The **$p$-adic absolute value** is \begin{align*} |x|_p := p^{-k}, \qquad |0|_p := 0. \end{align*} This measures "divisibility by $p$": the more times $p$ divides $x$, the *smaller* $|x|_p$ is. For example, $|p^{10}|_p = p^{-10}$ is tiny, while $|1/p^{10}|_p = p^{10}$ is large. The $p$-adic absolute value satisfies a property *stronger* than the triangle inequality: \begin{align*} |x + y|_p \leq \max(|x|_p, |y|_p), \end{align*} called the **ultrametric inequality**. This is strictly stronger than $|x+y|_p \leq |x|_p + |y|_p$ and is the hallmark of a non-Archimedean absolute value. In any ultrametric space, the geometry is radically different from Euclidean geometry: every triangle is isosceles, every point inside a ball is a center, and disjoint balls have a definite gap between them. The $p$-adic numbers $\mathbb{Q}_p$ are the completion of $\mathbb{Q}$ with respect to $|\cdot|_p$. They form a complete metric space that is non-Archimedean in the following sense: $|n|_p \leq 1$ for every $n \in \mathbb{N}$, since natural numbers have non-negative $p$-adic valuation ($n = p^k \cdot m$ with $k \geq 0$, so $|n|_p = p^{-k} \leq 1$). The sequence $1, 2, 3, \ldots$ is therefore *bounded* in $\mathbb{Q}_p$ — a stark contrast to the real numbers, where this sequence is unbounded. The natural numbers do not "march off to infinity" in $\mathbb{Q}_p$; instead, they remain confined to the closed unit ball $\{x \in \mathbb{Q}_p : |x|_p \leq 1\}$. [/explanation] The $p$-adic numbers illustrate that non-Archimedean behavior is not a pathology but a genuinely different (and mathematically rich) way of measuring size. The Archimedean property is really a property of the *absolute value* $|\cdot|$, not just the ordering. Ostrowski's theorem classifies all nontrivial absolute values on $\mathbb{Q}$: they are either the usual absolute value (Archimedean) or a $p$-adic absolute value for some prime $p$ (non-Archimedean). In this sense, the Archimedean property singles out the familiar real numbers among all completions of $\mathbb{Q}$. ## Archimedean Ordered Groups The Archimedean property extends naturally beyond fields to ordered groups, where it becomes a structural condition with deep consequences. [definition: Archimedean Ordered Group] A totally ordered abelian group $(G, +, <)$ is **Archimedean** if for every $g, h \in G$ with $g > 0$, there exists $n \in \mathbb{N}$ such that $ng > h$ (where $ng = \underbrace{g + g + \cdots + g}_{n}$). [/definition] The condition says that no element is "infinitely larger" than any other positive element — every positive element, repeated enough times, exceeds any given element. This is the original formulation due to Archimedes (phrased for lengths): given any two line segments, sufficiently many copies of the shorter placed end-to-end exceed the longer. The Archimedean property for ordered groups has a striking structural consequence: it forces the group to be "essentially one-dimensional." [quotetheorem:1236] Holder's theorem says that every Archimedean ordered group is isomorphic (as an ordered group) to a subgroup of $(\mathbb{R}, +)$. Since the subgroups of $(\mathbb{R}, +)$ are either dense in $\mathbb{R}$ or cyclic (of the form $a\mathbb{Z}$ for some $a \geq 0$), this completely classifies Archimedean ordered groups: they are either cyclic or dense subgroups of $\mathbb{R}$. This classification has a notable consequence. The value group of a non-Archimedean absolute value (such as $|\cdot|_p$) is the group $\{p^k : k \in \mathbb{Z}\} \cong \mathbb{Z}$ under multiplication, which *is* Archimedean as an ordered group. The non-Archimedean behavior of $|\cdot|_p$ comes from the ultrametric inequality, not from the value group. This shows that "Archimedean" means different things in different contexts: for ordered *fields*, it concerns the unboundedness of $\mathbb{N}$; for *absolute values*, it concerns the triangle inequality; for ordered *groups*, it concerns the absence of infinite elements. ## Standard Techniques The Archimedean property is so deeply embedded in analysis that it functions as a proof technique rather than a theorem to be cited. The following are the standard patterns. ### Approximation by Rationals **Pattern:** To prove a property holds for all $x \in \mathbb{R}$, first prove it for $x \in \mathbb{Q}$, then extend by density. This technique requires: 1. The density of $\mathbb{Q}$ in $\mathbb{R}$ (which depends on the Archimedean property). 2. The property in question must be preserved under limits (continuity, or a closure argument). [example: Continuous Functions Determined By Rational Values] Let $f, g: \mathbb{R} \to \mathbb{R}$ be continuous functions with $f(r) = g(r)$ for all $r \in \mathbb{Q}$. Then $f = g$ on all of $\mathbb{R}$. To see this, fix $x \in \mathbb{R}$. By the density of $\mathbb{Q}$, for each $n \in \mathbb{N}$, there exists $r_n \in \mathbb{Q}$ with $|r_n - x| < 1/n$ (here we use the Archimedean property to find $n$ with $1/n$ smaller than any desired tolerance). Then $r_n \to x$, and by continuity, \begin{align*} f(x) = \lim_{n \to \infty} f(r_n) = \lim_{n \to \infty} g(r_n) = g(x). \end{align*} This argument fails in non-Archimedean settings: in the hyperreals, continuous functions are not determined by their values on $\mathbb{Q}$ because $\mathbb{Q}$ is not dense (there are infinitesimal gaps). [/example] ### The $1/n$ Squeeze **Pattern:** To show $a \leq b$ when a direct comparison is difficult, show $a \leq b + 1/n$ for every $n \in \mathbb{N}$, then take $n \to \infty$. This technique rests on the Archimedean property in the following precise sense: if $a - b > 0$, then by the Archimedean property there exists $n$ with $1/n < a - b$, contradicting $a \leq b + 1/n$. The contrapositive gives $a \leq b$. [example: Squeeze For Infimum Characterization] Let $A \subset \mathbb{R}$ be nonempty and bounded below. To show that $\alpha = \inf A$ satisfies a property $P(\alpha)$, a common strategy is: 1. For each $n \in \mathbb{N}$, choose $a_n \in A$ with $\alpha \leq a_n < \alpha + 1/n$ (possible because $\alpha + 1/n$ is not a lower bound for $A$, since $\alpha$ is the *greatest* lower bound). 2. Then $a_n \to \alpha$ by the squeeze theorem (since $|a_n - \alpha| < 1/n \to 0$ by the Archimedean property). 3. If $P$ is preserved under limits, conclude $P(\alpha)$. Step 1 uses the characterization of infimum. Step 2 uses the Archimedean property. This pattern appears in the proofs of the intermediate value theorem, the extreme value theorem, and throughout measure theory. [/example] ### The Archimedean Dichotomy **Pattern:** Given $x \in \mathbb{R}$ with $x \geq 0$, either $x = 0$ or $x > 0$. If $x > 0$, the Archimedean property produces useful consequences; if $x = 0$, the conclusion often follows directly. More precisely: if $x \geq 0$ and $x < 1/n$ for all $n \in \mathbb{N}$, then $x = 0$. This is the Archimedean property in its role as a "no infinitesimals" guarantee, and it is the mechanism behind proofs like: "if $|a - b| < \varepsilon$ for all $\varepsilon > 0$, then $a = b$." [example: Uniqueness Of Limits] The uniqueness of limits for convergent sequences uses the Archimedean dichotomy. Suppose $a_n \to L$ and $a_n \to M$. Then for every $\varepsilon > 0$, there exists $N$ such that $|a_n - L| < \varepsilon/2$ and $|a_n - M| < \varepsilon/2$ for $n \geq N$. By the triangle inequality, $|L - M| < \varepsilon$. Since this holds for every $\varepsilon > 0$, and $|L - M| \geq 0$, the Archimedean property forces $|L - M| = 0$, so $L = M$. In a non-Archimedean field, this argument would fail: $|L - M|$ could be a positive infinitesimal, less than every standard $\varepsilon > 0$ but not zero. Limits would not be unique. [/example] ## References - Rudin, W., *Principles of Mathematical Analysis*, 3rd ed. (1976), Chapter 1. - Bartle, R. and Sherbert, D., *Introduction to Real Analysis*, 4th ed. (2011), Section 2.4. - Schechter, E., *Handbook of Analysis and Its Foundations* (1997), Sections 10.5--10.7. - Robert, A., *A Course in $p$-adic Analysis* (2000), Chapter 1. - Goldblatt, R., *Lectures on the Hyperreals: An Introduction to Nonstandard Analysis* (1998), Chapters 1--3.