# Completeness of the Real Numbers and Cauchy Sequences: The Absence of Gaps

> From the axiom that every bounded increasing sequence converges we derive the Archimedean property, suprema, density, Bolzano-Weierstrass and Cauchy completeness, omitting no steps.
> https://rikai.mugen-giken.com/en/mathematics/real-analysis/completeness-and-cauchy

## 0. Key points

- The field $\mathbb{R}$ of real numbers is an ordered field together with one extra requirement: **completeness**. The field and order axioms alone cannot distinguish $\mathbb{R}$ from the rational field $\mathbb{Q}$, and they cannot even prove that $\sqrt{2}$ exists.
- Here we adopt completeness in the form of the axiom "**every increasing sequence that is bounded above converges**", and from it we derive, without skipping any step, the Archimedean property, the existence of suprema, the density of the rationals and of the irrationals, the Bolzano–Weierstrass theorem, and the convergence of Cauchy sequences.
- The Archimedean property (every real number is smaller than some natural number) does not follow from the ordered-field axioms. It is a consequence of completeness, and it can be read as the statement that "there are no infinitely small numbers".
- Convergence of Cauchy sequences means that **convergence can be decided without naming the limit**. This is the practical watershed separating calculus as a computational technique from real analysis as a study of foundations.
- The completeness established here reappears, in changed form, again and again: in uniform convergence, in the convergence theorems of Lebesgue integration, and in the completeness of $L^p$ spaces (the Riesz–Fischer theorem).

## 1. Motivation: what calculus used silently

In calculus we learned the rules for computing limits, derivatives and integrals. Along the way we accepted the following facts without proof, or on the strength of a picture.

- The number $\sqrt{2}$ exists.
- If a continuous function changes sign, it takes the value $0$ somewhere in between (the intermediate value theorem).
- A continuous function on a bounded closed interval attains a maximum.
- An increasing sequence that is bounded above converges.

Every one of these depends on the single fact that the real line has no gaps. This becomes plain once we move the stage to the world $\mathbb{Q}$ of rational numbers. The function $f(x) = x^2 - 2$ is continuous on $\mathbb{Q}$, with $f(0) = -2$ negative and $f(2) = 2$ positive. Nevertheless no rational $q$ satisfies $f(q) = 0$. The intermediate value theorem fails over $\mathbb{Q}$ (<Ref to="mathematics/real-analysis/uniform-continuity#rem-rational-ivt" />). The extreme value theorem breaks in the same way. The function $g(x) = \dfrac{1}{x^2 - 2}$ is continuous on the bounded closed interval $[0, 2] \cap \mathbb{Q}$ of $\mathbb{Q}$, yet it is not even bounded, because $x^2$ can be brought arbitrarily close to $2$.

Neither the rules of arithmetic (the field axioms) nor the rules governing inequalities (the order axioms) distinguish $\mathbb{Q}$ from $\mathbb{R}$: both are ordered fields. So the four facts above **can never be derived from arithmetic and order alone**, and one further requirement must be imposed. That requirement is completeness.

Historically this realisation came late. Analysis up to the eighteenth century leaned on an intuition of "magnitude", and the question of what a real number is was not asked. Bolzano (1817) attempted a "purely analytic proof" of the intermediate value theorem, and Cauchy's *Cours d'analyse* (1821) rebuilt calculus around the notion of limit; but neither supplied a definition of the real numbers themselves. In 1872 two constructions appeared — Dedekind's "cuts" and Cantor's (and Méray's) "fundamental sequences" — and only then were the reals defined. Real analysis is the discipline of rederiving from axioms what calculus took for granted. This article is its first step.

<Aside type="note">
In what follows we assume familiarity with $\varepsilon$-$N$ and $\varepsilon$-$\delta$ arguments and with the handling of quantifiers. If you are unsure, read [Limits and continuity (the ε-δ argument)](/en/mathematics/calculus/limits-and-continuity) first (in particular <Ref to="mathematics/calculus/limits-and-continuity#def-limit" />) together with [The grammar of mathematics — sets and logic](/en/mathematics/foundations/sets-and-logic).
</Aside>

## 2. Preliminaries: ordered fields and convergence of sequences

Let us first record, as axioms, the "arithmetic and order" part, that is, everything except completeness. Making this explicit lets us say later, with precision, that a given statement does not follow from the ordered-field axioms alone.

<Definition id="def-ordered-field" title="Ordered field">
Let a set $K$ carry an addition $+$, a multiplication $\cdot$ and a total order $\le$ satisfying the following three conditions. Then $(K, +, \cdot, \le)$ is called an **ordered field**.

1. $(K, +, \cdot)$ is a field: addition and multiplication are associative and commutative, there is an additive identity $0$ and a multiplicative identity $1 \ne 0$, every element $a$ has an additive inverse $-a$, every element $a \ne 0$ has a multiplicative inverse $a^{-1}$, and the distributive law $a(b + c) = ab + ac$ holds.
2. For all $a, b, c \in K$, if $a \le b$ then $a + c \le b + c$.
3. For all $a, b \in K$, if $0 \le a$ and $0 \le b$ then $0 \le ab$.
</Definition>

Both $\mathbb{Q}$ and $\mathbb{R}$ are ordered fields. In any ordered field $0 < 1$ follows. Indeed, $a^2 \ge 0$ for every $a$: if $a \ge 0$ this is immediate from condition 3, while if $a \le 0$ we add $-a$ to both sides using condition 2 to get $-a \ge 0$ and then apply condition 3 to $(-a)^2 = a^2$. Hence $1 = 1^2 \ge 0$, and together with $1 \ne 0$ this gives $0 < 1$. Consequently $1,\ 1 + 1,\ 1 + 1 + 1,\ \ldots$ is, by condition 2, a strictly increasing sequence of positive elements, each step adding $1$. In particular these elements are pairwise distinct, so **every ordered field contains a copy of the natural numbers**, and with it copies of the integers and of the rationals. Throughout, $\mathbb{N} = \{1, 2, 3, \ldots\}$ does not include $0$.

We define the absolute value by $|a| = \max\{a, -a\}$. Adding the inequalities $a \le |a|$, $-a \le |a|$, $b \le |b|$ and $-b \le |b|$ in pairs gives $a + b \le |a| + |b|$ and $-(a + b) \le |a| + |b|$, hence the triangle inequality $|a + b| \le |a| + |b|$. This inequality is used repeatedly in what follows.

<Definition id="def-convergence" title="Convergence, boundedness and monotonicity of sequences">
Let $(a_n)_{n \ge 1}$ be a sequence of elements of an ordered field $K$.

- We say that it **converges** to $L \in K$ if $\forall \varepsilon > 0,\ \exists N \in \mathbb{N},\ \forall n \ge N,\ |a_n - L| < \varepsilon$, and we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$.
- It is **bounded** if there is an $M > 0$ with $|a_n| \le M$ for every $n$.
- It is **increasing** if $a_n \le a_{n+1}$ for every $n$. It is **decreasing** if the inequality is reversed.
</Definition>

<Remark id="rem-uniqueness-of-limit">
The limit is unique. Indeed, suppose $a_n \to L$ and $a_n \to L'$ with $L \ne L'$. Taking $\varepsilon = |L - L'| / 2 > 0$, we have $|a_n - L| < \varepsilon$ and $|a_n - L'| < \varepsilon$ for all sufficiently large $n$, so the triangle inequality gives $|L - L'| \le |L - a_n| + |a_n - L'| < 2\varepsilon = |L - L'|$, a contradiction. This uniqueness is exactly what we shall need later when we assert that a certain limit "does not exist inside $\mathbb{Q}$".
</Remark>

## 3. The completeness axiom and the Archimedean property

There are several equivalent ways to formulate completeness (we collect them after <Ref to="thm-supremum" /> and in the Appendix at the end of the article). Here we take as our axiom the form most familiar from calculus.

<Axiom id="ax-completeness" title="Completeness axiom (monotone convergence principle)">
Let $(a_n)_{n \ge 1}$ be a sequence of real numbers that is increasing and bounded above, that is, $a_n \le a_{n+1}$ for every $n$ and there exists $M \in \mathbb{R}$ with $a_n \le M$ for every $n$. Then $(a_n)$ converges to some real number.
</Axiom>

What this axiom asserts is that a destination has been provided. The sequence $1,\ 1.4,\ 1.41,\ 1.414,\ \ldots$ is increasing and never exceeds $2$, but inside $\mathbb{Q}$ there is nowhere for it to go. Inside $\mathbb{R}$ there is: that is the content of the axiom. From now on, $\mathbb{R}$ means an ordered field satisfying this axiom.

The first consequence is the Archimedean property. The name sounds elementary, but note that it does **not** follow from the ordered-field axioms.

<Theorem id="thm-archimedes" title="Archimedean property">
The following hold.

1. For every $x \in \mathbb{R}$ there exists $n \in \mathbb{N}$ with $n > x$.
2. For every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ with $1/N < \varepsilon$. In particular $\lim_{n \to \infty} 1/n = 0$.
3. $\lim_{n \to \infty} 2^{-n} = 0$.
</Theorem>

<Proof of="thm-archimedes">
(1) Suppose the conclusion fails, so that there is an $x \in \mathbb{R}$ with $n \le x$ for every $n \in \mathbb{N}$. The sequence $a_n = n$ is increasing and, by this assumption, has $x$ as an upper bound. Hence by <Ref to="ax-completeness" /> the sequence $(a_n)$ converges to some real number $L$. Taking $\varepsilon = 1/2$ in <Ref to="def-convergence" />, there is an $N$ such that $|a_n - L| < 1/2$ for all $n \ge N$. Applying this to $n = N$ and $n = N + 1$ and using the triangle inequality,
$$
1 = |a_{N+1} - a_N| \le |a_{N+1} - L| + |L - a_N| < \tfrac12 + \tfrac12 = 1
$$
which gives the contradiction $1 < 1$. Hence (1) holds.

(2) Let $\varepsilon > 0$. Apply (1) with $x = 1/\varepsilon$ to obtain $N \in \mathbb{N}$ with $N > 1/\varepsilon$. Multiplying both sides of $N > 1/\varepsilon > 0$ by the positive number $\varepsilon / N$ gives $\varepsilon > 1/N$. Moreover, if $n \ge N$ then $0 < 1/n \le 1/N < \varepsilon$, so $|1/n - 0| < \varepsilon$, which proves $1/n \to 0$.

(3) We have $2^n \ge n$ for every $n \in \mathbb{N}$. Indeed, $2 \ge 1$ for $n = 1$, and assuming $2^n \ge n$ we get $2^{n+1} = 2 \cdot 2^n \ge 2n = n + n \ge n + 1$, so the claim follows by induction. Hence $0 < 2^{-n} \le 1/n$, and (2) together with the squeeze gives $2^{-n} \to 0$. Spelling out the squeeze: given $\varepsilon > 0$, take the $N$ from (2); then $|2^{-n} - 0| = 2^{-n} \le 1/n < \varepsilon$ for $n \ge N$.
</Proof>

<Example id="ex-non-archimedean" title="An ordered field in which the Archimedean property fails">
That the Archimedean property is not a consequence of the ordered-field axioms can be checked by a counterexample. The set of all rational functions with real coefficients,
$$
\mathbb{R}(t) = \left\{ \frac{p(t)}{q(t)} \ \middle|\ p, q \text{ real polynomials},\ q \ne 0 \right\}
$$
is a field. Order it by declaring that $f > 0$ means $f(s) > 0$ for all sufficiently large real $s$. A rational function that is not identically $0$ has a constant sign for all large $s$, so this does define a total order, and conditions 2 and 3 of <Ref to="def-ordered-field" /> hold as well. In this ordered field, $t - n$ is positive for all large $s$ for every $n \in \mathbb{N}$, so $t > n$. Thus the element $t$ exceeds every natural number and the Archimedean property fails. Its reciprocal $1/t$ is an "infinitesimal": positive, yet smaller than every $1/n$.

In this sense the Archimedean property states that there are no infinitely large and no infinitely small elements, and what guarantees it is the completeness axiom. That is the content of <Ref to="thm-archimedes" />.
</Example>

Before proving anything further, here is a map of how the statements ahead depend on one another.

<Figure caption="The chain of reasoning starting from the completeness axiom. An arrow means 'is used to prove'">
<Mermaid code={`flowchart TD
  A["Axiom: an increasing sequence bounded above converges"] --> B["Archimedean property"]
  C["Monotone subsequence lemma"]
  A --> D["Existence of suprema"]
  B --> D
  D --> E["Existence of square roots"]
  B --> F["Existence of the integer part"]
  F --> G["Density of the rationals"]
  G --> H["Density of the irrationals"]
  E --> H
  C --> I["Bolzano-Weierstrass theorem"]
  A --> I
  I --> J["Cauchy sequences converge"]`} />
</Figure>

## 4. Existence of suprema and the construction of square roots

<Definition id="def-supremum" title="Upper bound, supremum, infimum">
Let $S \subset \mathbb{R}$.

- An element $b \in \mathbb{R}$ is an **upper bound** of $S$ if $x \le b$ for every $x \in S$. If an upper bound exists, $S$ is said to be **bounded above**.
- An element $L \in \mathbb{R}$ is a **supremum** of $S$ if $L$ is an upper bound of $S$ and $L \le b$ for every upper bound $b$ of $S$; that is, the supremum is the least upper bound. If it exists it is unique (two of them would each be at most the other), and we write it $\sup S$.
- **Lower bound**, **bounded below** and **infimum** $\inf S$ are defined in the same way with the inequalities reversed.
</Definition>

The supremum differs from a maximum in that it need not belong to $S$. The set $S = \{x \in \mathbb{R} \mid 0 < x < 1\}$ has no maximum, yet $\sup S = 1$. The guarantee that a least upper bound exists even when a maximum does not is precisely the next theorem.

<Theorem id="thm-supremum" title="Existence of suprema (Weierstrass)">
If $S \subset \mathbb{R}$ is non-empty and bounded above, then $\sup S$ exists. Likewise, if $S$ is non-empty and bounded below, then $\inf S$ exists.
</Theorem>

<Proof of="thm-supremum">
We build a sequence converging to the supremum by repeatedly halving an interval (bisection).

**Construction of the sequences.** Below we index from $n = 0$; shifting everything by $1$ turns the sequences into sequences of the form used in <Ref to="def-convergence" />, so <Ref to="ax-completeness" /> applies unchanged. Since $S \ne \emptyset$, pick some $s_0 \in S$ and set $a_0 = s_0 - 1$. As $s_0 \in S$ and $s_0 > a_0$, the number $a_0$ is not an upper bound of $S$. By hypothesis an upper bound $b_0$ exists, and $a_0 < s_0 \le b_0$ gives $a_0 < b_0$.

Now suppose $a_n < b_n$, that $a_n$ is not an upper bound of $S$ and that $b_n$ is. Take the midpoint $m_n = (a_n + b_n)/2$ and set

- $a_{n+1} = a_n$, $b_{n+1} = m_n$ if $m_n$ is an upper bound of $S$;
- $a_{n+1} = m_n$, $b_{n+1} = b_n$ if $m_n$ is not an upper bound of $S$.

Since $a_n < b_n$ implies $a_n < m_n < b_n$, in either case $a_{n+1} < b_{n+1}$, with $a_{n+1}$ not an upper bound and $b_{n+1}$ an upper bound. Moreover $a_n \le a_{n+1}$ and $b_{n+1} \le b_n$, and the widths satisfy
$$
b_{n+1} - a_{n+1} = \frac{b_n - a_n}{2}, \qquad \text{hence} \quad b_n - a_n = \frac{b_0 - a_0}{2^{n}}
$$
(the second identity by induction on $n$).

**Existence of the limit.** The sequence $(a_n)$ is increasing and $a_n < b_n \le b_0$ for every $n$, so it is bounded above. By <Ref to="ax-completeness" /> there is an $L \in \mathbb{R}$ with $a_n \to L$. We show next that $b_n \to L$. Let $\varepsilon > 0$. Apply <Ref to="thm-archimedes" />(3) with $\varepsilon' = \dfrac{\varepsilon}{2(b_0 - a_0)} > 0$ to obtain $N_1$ with $2^{-n} < \varepsilon'$ for $n \ge N_1$; then $b_n - a_n = (b_0 - a_0)2^{-n} < \varepsilon/2$ for $n \ge N_1$. Also, since $a_n \to L$, there is $N_2$ with $|a_n - L| < \varepsilon/2$ for $n \ge N_2$. For $n \ge \max\{N_1, N_2\}$ the triangle inequality gives
$$
|b_n - L| \le |b_n - a_n| + |a_n - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
$$
so $b_n \to L$.

**$L$ is the supremum.** First, $L$ is an upper bound. Otherwise there is an $x \in S$ with $x > L$. Applying $b_n \to L$ with $\varepsilon = x - L > 0$, we find an $n$ with $|b_n - L| < x - L$ and hence $b_n < L + (x - L) = x$, contradicting the fact that $b_n$ is an upper bound of $S$. So $L$ is an upper bound.

Second, $L$ is the least upper bound. It suffices to show that no $c < L$ is an upper bound. Applying $a_n \to L$ with $\varepsilon = L - c > 0$, we find an $n$ with $|a_n - L| < L - c$ and hence $a_n > L - (L - c) = c$. This $a_n$ is not an upper bound of $S$, so some $x \in S$ satisfies $x > a_n > c$. Therefore $c$ is not an upper bound. Together this gives $L = \sup S$.

**The infimum.** If $S$ is non-empty and bounded below, then $-S = \{-x \mid x \in S\}$ is non-empty and bounded above (if $b$ is a lower bound of $S$ then $-b$ is an upper bound of $-S$), so $\sup(-S)$ exists by what we have just proved. That $\inf S = -\sup(-S)$ follows at once from the equivalence $x \ge c \iff -x \le -c$.
</Proof>

<Ref to="thm-supremum" /> is often adopted itself as the completeness axiom, under the name "least upper bound property". Here we derived it from <Ref to="ax-completeness" />. Conversely, <Ref to="ax-completeness" /> can be derived from <Ref to="thm-supremum" /> (use the characterisation in <Ref to="exr-sup-characterization" /> and check that the supremum of the range of an increasing sequence $(a_n)$ is its limit). The two are equivalent.

Let us test the power of this theorem by constructing $\sqrt{2}$. That "a square root of $2$ exists" is something to be proved from the definition of $\mathbb{R}$.

<Proposition id="prop-sqrt-two" title="Existence of positive square roots">
There is exactly one real number $\alpha$ with $\alpha^2 = 2$ and $\alpha > 0$.
</Proposition>

<Proof of="prop-sqrt-two">
Set $S = \{x \in \mathbb{R} \mid x > 0,\ x^2 < 2\}$. Since $1 > 0$ and $1^2 = 1 < 2$ we have $1 \in S$, so $S \ne \emptyset$. Also, if $x > 2$ then $x^2 > 2x > 4 > 2$ and hence $x \notin S$; thus every element of $S$ is at most $2$, i.e. $2$ is an upper bound. By <Ref to="thm-supremum" /> the number $\alpha = \sup S$ exists, and $1 \in S$ gives $\alpha \ge 1 > 0$.

We now show that $\alpha^2 < 2$ and $\alpha^2 > 2$ are both impossible. As preparation, note that for positive $u, v$ we have $u^2 < v^2 \implies u < v$ (contrapositive: if $u \ge v > 0$ then $u^2 \ge uv \ge v^2$).

**Case $\alpha^2 < 2$.** By <Ref to="thm-archimedes" />(2) we may choose $n \in \mathbb{N}$ with
$$
\frac{1}{n} < \min\left\{ 1,\ \frac{2 - \alpha^2}{2\alpha + 1} \right\}
$$
(the right-hand side is positive). Put $h = 1/n$. From $0 < h < 1$ we get $h^2 < h$, whence
$$
(\alpha + h)^2 = \alpha^2 + 2\alpha h + h^2 < \alpha^2 + 2\alpha h + h = \alpha^2 + (2\alpha + 1)h < \alpha^2 + (2 - \alpha^2) = 2 .
$$
Since also $\alpha + h > 0$, we have $\alpha + h \in S$; as $\alpha$ is an upper bound this forces $\alpha + h \le \alpha$, i.e. $h \le 0$, contradicting $h > 0$.

**Case $\alpha^2 > 2$.** Again by <Ref to="thm-archimedes" />(2), choose $n$ with
$$
\frac{1}{n} < \min\left\{ \alpha,\ \frac{\alpha^2 - 2}{2\alpha} \right\}
$$
and put $k = 1/n$. From $0 < k < \alpha$ we get $\alpha - k > 0$, and
$$
(\alpha - k)^2 = \alpha^2 - 2\alpha k + k^2 > \alpha^2 - 2\alpha k > \alpha^2 - (\alpha^2 - 2) = 2 .
$$
Then every $x \in S$ satisfies $x^2 < 2 < (\alpha - k)^2$, and since $x > 0$ and $\alpha - k > 0$ the remark above gives $x < \alpha - k$. So $\alpha - k$ is an upper bound of $S$, and $\alpha - k < \alpha$ contradicts the fact that $\alpha$ is the least upper bound.

Hence $\alpha^2 = 2$. For uniqueness, if $\beta > 0$ satisfies $\beta^2 = 2$ then $(\alpha - \beta)(\alpha + \beta) = \alpha^2 - \beta^2 = 0$, and $\alpha + \beta > 0$ forces $\alpha = \beta$. We write this $\alpha$ as $\sqrt{2}$.
</Proof>

That this $\sqrt{2}$ is irrational follows from the familiar argument by contradiction. If $\sqrt{2} = p/q$ with $p, q$ coprime natural numbers, then $p^2 = 2q^2$, so $p$ is even; writing $p = 2p'$ gives $4p'^2 = 2q^2$, hence $q^2 = 2p'^2$ and $q$ is even too, contradicting coprimality. (For arguments of this type see <Ref to="mathematics/foundations/proof-techniques#thm-sqrt2" /> in [Proof techniques — induction and contradiction](/en/mathematics/foundations/proof-techniques).) Combining these two results, we can say precisely in what way $\mathbb{Q}$ fails to be complete.

<Example id="ex-q-no-sup" title="A set with no supremum in the world of rational numbers">
Consider $S_{\mathbb{Q}} = \{x \in \mathbb{Q} \mid x > 0,\ x^2 < 2\}$. It is non-empty, since $1 \in S_{\mathbb{Q}}$, and $2$ is an upper bound. Nevertheless $S_{\mathbb{Q}}$ has **no supremum inside $\mathbb{Q}$**.

Here is why. Suppose $q \in \mathbb{Q}$ were the least upper bound of $S_{\mathbb{Q}}$ within $\mathbb{Q}$. From $1 \in S_{\mathbb{Q}}$ we get $q \ge 1 > 0$. Now repeat the proof of <Ref to="prop-sqrt-two" /> verbatim. The Archimedean property does hold in $\mathbb{Q}$ (for a rational $p/q'$ with $q' \ge 1$ we have $p/q' \le |p| < |p| + 1$, which follows directly from the ordered-field axioms alone). Hence the increments $h = 1/n$ and $k = 1/n$ can be chosen **rational** in the same way: if $q^2 < 2$ then $q + h \in S_{\mathbb{Q}}$, contradicting that $q$ is an upper bound, and if $q^2 > 2$ then the rational $q - k$ is an upper bound, contradicting minimality. The only remaining possibility is $q^2 = 2$, which contradicts the irrationality of $\sqrt{2}$.

So <Ref to="thm-supremum" /> fails in $\mathbb{Q}$ (<Ref to="mathematics/foundations/what-is-a-number#thm-q-has-gap" />). Although $\mathbb{Q}$ is an ordered field and satisfies the Archimedean property, it is not complete. What is still missing once arithmetic, order and the Archimedean property are all in place is exactly completeness.
</Example>

## 5. Density of the rationals and the irrationals

If completeness says "there are no gaps", we naturally want to know how the gaps are filled as seen from the rational side. That is density. First we construct integer parts.

<Lemma id="lem-integer-part" title="Existence of the integer part">
For every $x \in \mathbb{R}$ there is exactly one integer $m$ with $m \le x < m + 1$. We write it $\lfloor x \rfloor$.
</Lemma>

<Proof of="lem-integer-part">
Set $B = \{k \in \mathbb{Z} \mid k > x\}$. By <Ref to="thm-archimedes" />(1) there is $n_1 \in \mathbb{N}$ with $n_1 > x$, so $B \ne \emptyset$. Applying (1) again to $-x$, take $n_2 \in \mathbb{N}$ with $n_2 > -x$; then every $k \in B$ satisfies $k > x > -n_2$, so $B$ is bounded below by $-n_2$. A non-empty set of integers that is bounded below has a least element (the set $\{k + n_2 + 1 \mid k \in B\}$ is a non-empty subset of the natural numbers, hence has a least element by well-ordering; on well-ordering see <Ref to="mathematics/foundations/proof-techniques#ax-well-ordering" />).

Let $m_0 = \min B$ be that least element and put $m = m_0 - 1$. Then $m$ is an integer, and from $m < m_0$ and the minimality of $m_0$ we get $m \notin B$, i.e. $m \le x$. On the other hand $m + 1 = m_0 \in B$ gives $x < m + 1$.

For uniqueness, suppose integers $m \ne m'$ both satisfy $m \le x < m + 1$ and $m' \le x < m' + 1$; we may assume $m < m'$. As these are integers, $m + 1 \le m'$, and combined with $m' \le x$ this yields $m + 1 \le x$, contradicting $x < m + 1$.
</Proof>

<Theorem id="thm-density-rational" title="Density of the rationals">
If $a, b \in \mathbb{R}$ satisfy $a < b$, then there is a rational number $q$ with $a < q < b$.
</Theorem>

<Proof of="thm-density-rational">
Since $b - a > 0$, <Ref to="thm-archimedes" />(2) provides $n \in \mathbb{N}$ with $\dfrac{1}{n} < b - a$, i.e. $n(b - a) > 1$. For this $n$, apply <Ref to="lem-integer-part" /> to $na$ and put $m = \lfloor na \rfloor + 1$. By definition
$$
m - 1 \le na < m, \qquad \text{that is} \qquad na < m \le na + 1 .
$$
Put $q = m/n$. Since $n > 0$, the left inequality $na < m$ gives $a < m/n = q$. From the right inequality together with $n(b - a) > 1$,
$$
m \le na + 1 < na + n(b - a) = nb
$$
so dividing by $n > 0$ gives $q = m/n < b$. As $m$ is an integer and $n$ a natural number, $q$ is rational, and $a < q < b$ is proved.
</Proof>

<Corollary id="cor-density-irrational" title="Density of the irrationals">
If $a, b \in \mathbb{R}$ satisfy $a < b$, then there is an irrational number $\xi$ (that is, $\xi \in \mathbb{R} \setminus \mathbb{Q}$) with $a < \xi < b$.
</Corollary>

<Proof of="cor-density-irrational">
We use the number $\sqrt{2}$ whose existence was guaranteed by <Ref to="prop-sqrt-two" />. Since $a - \sqrt{2} < b - \sqrt{2}$, <Ref to="thm-density-rational" /> provides a rational $q$ with
$$
a - \sqrt{2} < q < b - \sqrt{2} .
$$
Put $\xi = q + \sqrt{2}$; adding $\sqrt{2}$ to each term gives $a < \xi < b$. If $\xi$ were rational, then $\sqrt{2} = \xi - q$ would be rational too, contradicting the irrationality of $\sqrt{2}$. Hence $\xi$ is irrational.
</Proof>

<Example id="ex-density-computation" title="Producing a rational exactly as the proof of density prescribes">
The proof of <Ref to="thm-density-rational" /> is itself a recipe for producing a rational number. Let us construct one between $a = \sqrt{2} = 1.41421356\ldots$ and $b = \sqrt{2} + 10^{-3}$.

Since $b - a = 10^{-3}$, we take an $n$ with $1/n < 10^{-3}$, say $n = 1001$. Next we evaluate $na = 1001\sqrt{2}$:
$$
1001 \times 1.41421356\ldots = 1414.21356\ldots + 1.41421356\ldots = 1415.6277\ldots
$$
so $\lfloor na \rfloor = 1415$, hence $m = 1416$ and $q = \dfrac{1416}{1001}$.

Let us check. Since $1416 > 1415.6277\ldots = 1001a$, we have $q > a$. Also $1001b = 1001a + 1001 \times 10^{-3} = 1415.6277\ldots + 1.001 = 1416.6287\ldots > 1416$, so $q < b$. In decimal form $q = 1416/1001 = 1.4145854\ldots$, which indeed lies between $1.4142135\ldots$ and $1.4152135\ldots$.
</Example>

<Remark id="rem-density-vs-completeness">
Density and completeness are different notions. $\mathbb{Q}$ is dense in $\mathbb{R}$, yet as <Ref to="ex-q-no-sup" /> shows it is not complete. "Between any two points one can always squeeze in" is not the same as "there are no gaps". The gap between the two is also visible from the point of view of cardinality: $\mathbb{Q}$ is countable while $\mathbb{R}$ is uncountable, and by <Ref to="cor-density-irrational" /> the irrationals are dense as well. For how a countable dense set is scattered over an uncountable line, see [Cardinality and infinity — infinities come in sizes](/mathematics/foundations/cardinality-and-infinity). Incidentally, that the identity $0.999\ldots = 1$ is a theorem following from completeness rather than a notational convention is treated in <Ref to="mathematics/foundations/what-is-a-number#thm-0999" /> in [What is a number? (is 1 = 0.999...?)](/en/mathematics/foundations/what-is-a-number).
</Remark>

## 6. The Bolzano–Weierstrass theorem

We now turn to the existence theorem most used in analysis. A bounded sequence need not converge ($a_n = (-1)^n$ is a counterexample). But one can always extract a convergent part, and that is the next theorem, used over and over in proofs of the extreme value theorem and of uniform continuity.

<Definition id="def-subsequence" title="Subsequence">
Given a sequence $(a_n)_{n \ge 1}$ and a strictly increasing sequence of natural numbers $n_1 < n_2 < n_3 < \cdots$, the sequence $(a_{n_k})_{k \ge 1}$ is called a **subsequence** of $(a_n)$. Note that $n_1 \ge 1$ and $n_{k+1} > n_k$ give $n_k \ge k$ by induction.
</Definition>

The key to the proof is the following combinatorial lemma. What makes it striking is that it assumes no boundedness at all.

<Lemma id="lem-monotone-subsequence" title="Monotone subsequence lemma">
Every real sequence $(a_n)_{n \ge 1}$ has an increasing subsequence or a decreasing subsequence.
</Lemma>

<Proof of="lem-monotone-subsequence">
Call an index $n$ a **peak** if $a_n \ge a_m$ for every $m > n$. A peak is the position of a term that is no smaller than any later term. We distinguish cases according to the number of peaks.

**Infinitely many peaks.** List them in increasing order as $n_1 < n_2 < n_3 < \cdots$. For each $k$, the index $n_k$ is a peak and $n_{k+1} > n_k$, so by the definition of a peak $a_{n_k} \ge a_{n_{k+1}}$. Hence $(a_{n_k})$ is a decreasing subsequence.

**Only finitely many peaks.** If there are no peaks at all, put $N = 1$; otherwise let $N$ be one more than the largest peak. In either case no index $n \ge N$ is a peak. That $n \ge N$ is not a peak means, by negating the definition, that there exists $m > n$ with $a_m > a_n$. So put $n_1 = N$, and once $n_k$ is defined (it satisfies $n_k \ge N$ and so is not a peak) choose some $m > n_k$ with $a_m > a_{n_k}$ and set $n_{k+1} = m$. This produces $n_1 < n_2 < \cdots$ with $a_{n_1} < a_{n_2} < \cdots$, so $(a_{n_k})$ is an increasing subsequence.
</Proof>

<Figure caption="Peaks (in the accent colour) and the decreasing subsequence extracted from them. If there are infinitely many peaks, listing them already gives a decreasing sequence">
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</Figure>

<Theorem id="thm-bolzano-weierstrass" title="Bolzano–Weierstrass theorem">
Every bounded real sequence has a convergent subsequence. That is, if there is an $M > 0$ with $|a_n| \le M$ for every $n$, then there exist a subsequence $(a_{n_k})$ of $(a_n)$ and an $L \in \mathbb{R}$ with $a_{n_k} \to L$.
</Theorem>

<Proof of="thm-bolzano-weierstrass">
By <Ref to="lem-monotone-subsequence" />, the sequence $(a_n)$ has a monotone subsequence $(a_{n_k})$. Since $(a_n)$ is bounded, this subsequence satisfies $|a_{n_k}| \le M$ with the same $M$ and is therefore bounded.

If $(a_{n_k})$ is increasing, it is bounded above (by $a_{n_k} \le M$) and so converges to some $L$ by <Ref to="ax-completeness" />.

If $(a_{n_k})$ is decreasing, put $b_k = -a_{n_k}$. Then $(b_k)$ is increasing and bounded above by $b_k \le M$, so <Ref to="ax-completeness" /> gives an $L'$ with $b_k \to L'$. Setting $L = -L'$, for every $\varepsilon > 0$ and all sufficiently large $k$ we get
$$
|a_{n_k} - L| = |-b_k + L'| = |b_k - L'| < \varepsilon
$$
so $a_{n_k} \to L$. In both cases a convergent subsequence has been obtained.
</Proof>

<Example id="ex-alternating" title="Following the construction of the lemma on a sequence with no peaks">
Let $a_n = (-1)^n \dfrac{n}{n+1}$. Since $|a_n| < 1$ the sequence is bounded, but it does not converge, because the even-indexed terms approach $1$ and the odd-indexed terms approach $-1$.

This sequence has no peaks at all. Indeed, for every $n$ we have $a_n \le \dfrac{n}{n+1} < 1$, so $1 - a_n > 0$. By <Ref to="thm-archimedes" />(2) there is an $N$ with $\dfrac{1}{N} < 1 - a_n$, so we may choose an even number $2m$ large enough that $2m + 1 \ge N$ and $2m > n$ simultaneously. Then
$$
a_{2m} = \frac{2m}{2m+1} = 1 - \frac{1}{2m+1} \ge 1 - \frac{1}{N} > 1 - (1 - a_n) = a_n
$$
and since $2m > n$, the index $n$ is not a peak. As $n$ was arbitrary, there are no peaks.

We are therefore in the second case of <Ref to="lem-monotone-subsequence" />, and starting from $N = 1$ we can build an increasing subsequence. Following the recipe: for $n_1 = 1$ (with $a_1 = -\tfrac12$) the smallest $m$ with $a_m > -\tfrac12$ is $2$ (with $a_2 = \tfrac23$); the next $m$ with $a_m > \tfrac23$ is $4$ (with $a_4 = \tfrac45$); then $6$ (with $a_6 = \tfrac67$), and so on, giving
$$
-\frac{1}{2},\ \frac{2}{3},\ \frac{4}{5},\ \frac{6}{7},\ \frac{8}{9},\ \ldots \longrightarrow 1
$$
an increasing and bounded subsequence. By <Ref to="ax-completeness" /> it converges, and its limit is indeed $1$. What <Ref to="thm-bolzano-weierstrass" /> asserts is that this procedure always succeeds, for any bounded sequence.
</Example>

## 7. Cauchy sequences and the completeness of the reals

Both <Ref to="ax-completeness" /> and <Ref to="thm-bolzano-weierstrass" /> required structural information — monotonicity or boundedness — in order to conclude convergence. In practice, however, analysis deals with sequences such as $\sum_{k=1}^{n} \dfrac{(-1)^{k+1}}{k}$, which are neither monotone nor of known limit. The tool for deciding convergence without knowing the limit is Cauchy's condition.

<Definition id="def-cauchy" title="Cauchy sequence">
A real sequence $(a_n)_{n \ge 1}$ is a **Cauchy sequence** (a fundamental sequence) if
$$
\forall \varepsilon > 0,\ \exists N \in \mathbb{N},\ \forall m, n \ge N,\quad |a_n - a_m| < \varepsilon .
$$
</Definition>

<Aside type="caution">
Pay attention to the order and the range of the quantifiers. **Both** $m$ and $n$ range independently over the indices $\ge N$. The condition that consecutive terms get close, that is $|a_{n+1} - a_n| \to 0$, is much weaker. That the two are not equivalent is verified in <Ref to="exr-consecutive-differences" />.
</Aside>

<Proposition id="prop-cauchy-basic" title="Basic properties of Cauchy sequences">
For a real sequence $(a_n)$ the following hold.

1. If $(a_n)$ converges, then it is a Cauchy sequence.
2. If $(a_n)$ is a Cauchy sequence, then it is bounded.
</Proposition>

<Proof of="prop-cauchy-basic">
(1) Let $a_n \to L$ and let $\varepsilon > 0$ be arbitrary. Applying <Ref to="def-convergence" /> with $\varepsilon/2$, take $N$ with $|a_n - L| < \varepsilon/2$ for $n \ge N$. For $m, n \ge N$ the triangle inequality gives
$$
|a_n - a_m| \le |a_n - L| + |L - a_m| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
$$
so $(a_n)$ is a Cauchy sequence.

(2) Take $\varepsilon = 1$ in <Ref to="def-cauchy" /> and fix the corresponding $N$. For $n \ge N$, taking $m = N$ gives $|a_n - a_N| < 1$, and the triangle inequality yields
$$
|a_n| = |(a_n - a_N) + a_N| \le |a_n - a_N| + |a_N| < 1 + |a_N| .
$$
Now set $M = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|,\ |a_N| + 1\}$ (a maximum of finitely many real numbers, hence well defined). If $n < N$ then $|a_n| \le M$ by the definition of $M$; if $n \ge N$ then $|a_n| < |a_N| + 1 \le M$. In either case $|a_n| \le M$, so the sequence is bounded.
</Proof>

The converse — that every Cauchy sequence converges — is the heart of completeness, and it can be proved as a consequence of <Ref to="ax-completeness" />.

<Theorem id="thm-completeness" title="Completeness of the reals (Cauchy criterion)">
A real sequence $(a_n)_{n \ge 1}$ converges if and only if it is a Cauchy sequence.
</Theorem>

<Proof of="thm-completeness">
Necessity is <Ref to="prop-cauchy-basic" />(1). We prove sufficiency.

Let $(a_n)$ be a Cauchy sequence. By <Ref to="prop-cauchy-basic" />(2) it is bounded, so <Ref to="thm-bolzano-weierstrass" /> gives a convergent subsequence $(a_{n_k})$ with limit $L$. We show that the whole sequence $(a_n)$ converges to this $L$.

Let $\varepsilon > 0$ be arbitrary. First apply <Ref to="def-cauchy" /> with $\varepsilon/2$ to obtain $N_1$ with
$$
m, n \ge N_1 \implies |a_n - a_m| < \frac{\varepsilon}{2} .
$$
Next apply $a_{n_k} \to L$ with $\varepsilon/2$ to obtain $K$ with $|a_{n_k} - L| < \varepsilon/2$ for $k \ge K$.

Now put $k = \max\{K, N_1\}$. Then $k \ge K$, and $n_k \ge k$ (noted in <Ref to="def-subsequence" />) gives $n_k \ge N_1$. Fix this single index $n_k$. For any $n \ge N_1$ we have $n \ge N_1$ and $n_k \ge N_1$, so the Cauchy condition applies and
$$
|a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon .
$$
Since $\varepsilon > 0$ was arbitrary, $a_n \to L$.
</Proof>

The significance of this theorem is that **convergence can be decided without having a candidate for the limit**. Whether a sequence converges is settled purely by how close its terms are to one another, so the existence of a limit can be asserted before one knows what the limit is. Convergence tests for series, uniform convergence of sequences of functions, the construction of solutions of differential equations — most existence proofs in analysis have this shape.

<Example id="ex-newton-sqrt2" title="A Cauchy sequence that does not converge in the world of rational numbers">
Consider the sequence defined by $x_1 = 2$ and $x_{n+1} = \dfrac{x_n}{2} + \dfrac{1}{x_n}$ (this is Newton's method for $x^2 = 2$). It starts from a rational number and uses only rational arithmetic, so every $x_n$ is rational. Computing,
$$
x_1 = 2,\quad x_2 = \frac{3}{2},\quad x_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12},\quad x_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408} = 1.4142156\ldots
$$
which approaches $\sqrt{2} = 1.4142135\ldots$ very quickly.

That this converges in $\mathbb{R}$ follows from <Ref to="ax-completeness" />. First, $x_n > 0$ by induction (each term is built from positive numbers). Next, for $n \ge 1$,
$$
x_{n+1} - \sqrt{2} = \frac{x_n^2 - 2\sqrt{2}\,x_n + 2}{2x_n} = \frac{(x_n - \sqrt{2})^2}{2x_n} \ge 0
$$
so $x_n \ge \sqrt{2}$ for $n \ge 2$. Both sides being positive, multiplying the inequality by itself also gives $x_n^2 \ge 2$. Then
$$
x_n - x_{n+1} = \frac{x_n}{2} - \frac{1}{x_n} = \frac{x_n^2 - 2}{2x_n} \ge 0
$$
so $(x_n)_{n \ge 2}$ is decreasing with lower bound $\sqrt{2}$. Since $(-x_n)_{n \ge 2}$ is increasing with upper bound $-\sqrt{2}$, <Ref to="ax-completeness" /> applies and it converges, hence so does $(x_n)$. Writing $L$ for the limit, $x_n \ge \sqrt{2}$ for all $n \ge 2$ gives $L \ge \sqrt{2} > 0$. The sequence $(x_{n+1})$ is just $(x_n)$ with shifted indices, so it converges to the same $L$, and since $L \ne 0$ the limit law for quotients applies. Passing to the limit on both sides of the recursion gives $L = \dfrac{L}{2} + \dfrac{1}{L}$; multiplying by $2L$ and rearranging gives $L^2 = 2$, and $L > 0$ gives $L = \sqrt{2}$.

Now, since $(x_n)$ converges in $\mathbb{R}$, it is a Cauchy sequence by <Ref to="prop-cauchy-basic" />(1). Only rational numbers occur in the Cauchy condition $|x_n - x_m| < \varepsilon$, so this is also a "Cauchy sequence inside $\mathbb{Q}$". Yet by the uniqueness in <Ref to="rem-uniqueness-of-limit" /> the limit can only be $\sqrt{2}$, which is not rational. In other words, **in $\mathbb{Q}$ a Cauchy sequence need not converge**. <Ref to="thm-completeness" /> is a property specific to $\mathbb{R}$.
</Example>

<Example id="ex-harmonic" title="Proving divergence with the Cauchy condition">
Let $s_n = \displaystyle\sum_{k=1}^{n} \frac{1}{k}$. The sequence $(s_n)$ is increasing, but whether it is bounded above is not obvious at a glance. Here we use the negation of the Cauchy condition. Pairing the indices $n$ and $2n$: as $k$ runs from $n+1$ to $2n$ we have $k \le 2n$ and hence $\dfrac{1}{k} \ge \dfrac{1}{2n}$, and there are $n$ such terms, so
$$
|s_{2n} - s_n| = \sum_{k=n+1}^{2n} \frac{1}{k} \ge n \cdot \frac{1}{2n} = \frac{1}{2} .
$$
Therefore, for $\varepsilon = \dfrac{1}{2}$, whatever $N$ we choose the pair $n = N$, $m = 2N$ satisfies $|s_m - s_n| \ge \varepsilon$, so $(s_n)$ is not a Cauchy sequence. By the contrapositive of <Ref to="thm-completeness" />, $(s_n)$ does not converge.

Note that divergence was established without invoking any candidate for the limit. The argument used here is exactly the Cauchy criterion for series (<Ref to="mathematics/calculus/series-and-convergence#thm-cauchy-criterion" />). Convergence tests for series are treated in detail in [Series and convergence tests](/mathematics/calculus/series-and-convergence).
</Example>

<Remark id="rem-metric-completeness">
The only thing appearing in the definition of a Cauchy sequence is the "distance between two points" $|a_n - a_m|$. The notion therefore lifts to arbitrary metric spaces, and a metric space in which every Cauchy sequence converges is called a complete metric space. The completeness of $\mathbb{R}$ will reappear in disguise throughout what follows.

- The uniform Cauchy condition for sequences of functions guarantees uniform convergence without knowing the limit function (<Ref to="mathematics/real-analysis/uniform-convergence#thm-uniform-cauchy" />). See [Sequences of functions and uniform convergence](/mathematics/real-analysis/uniform-convergence).
- The convergence theorems of Lebesgue integration are a form of "completeness" that permits interchanging limits and integrals. They are treated in [Measurable sets and Lebesgue measure](/mathematics/real-analysis/lebesgue-measure) and [The Lebesgue integral and its convergence theorems](/mathematics/real-analysis/lebesgue-integral).
- The completeness of the $L^p$ spaces (the Riesz–Fischer theorem, <Ref to="mathematics/real-analysis/lp-spaces#thm-riesz-fischer" />) is the starting point of functional analysis. See [L^p spaces and an introduction to functional analysis](/mathematics/real-analysis/lp-spaces).

In the next chapter, [Continuous functions and uniform continuity](/en/mathematics/real-analysis/uniform-continuity), we shall see <Ref to="thm-bolzano-weierstrass" /> give birth to the extreme value theorem for continuous functions on bounded closed intervals and to uniform continuity (<Ref to="mathematics/real-analysis/uniform-continuity#thm-heine-cantor" />). Completeness remains the foundation there as well.
</Remark>

## 8. Exercises

<Exercise id="exr-sup-characterization" difficulty="Easy">
Let $S \subset \mathbb{R}$ be non-empty and bounded above, and let $L \in \mathbb{R}$. Show that $L = \sup S$ holds if and only if the following two conditions hold.

1. $x \le L$ for every $x \in S$.
2. For every $\varepsilon > 0$ there exists $x \in S$ with $x > L - \varepsilon$.

<Solution>
**Necessity.** Let $L = \sup S$. Condition 1 says that $L$ is an upper bound, which is the first half of <Ref to="def-supremum" />. We prove condition 2 by contradiction. Suppose that for some $\varepsilon_0 > 0$ there is no $x \in S$ with $x > L - \varepsilon_0$. Then every $x \in S$ satisfies $x \le L - \varepsilon_0$, so $L - \varepsilon_0$ is an upper bound of $S$. As $L$ is the least upper bound, $L \le L - \varepsilon_0$, i.e. $\varepsilon_0 \le 0$, contradicting $\varepsilon_0 > 0$.

**Sufficiency.** Assume conditions 1 and 2. By condition 1, $L$ is an upper bound. We show it is the least one. Let $b$ be any upper bound of $S$ and suppose $b < L$. Applying condition 2 with $\varepsilon = L - b > 0$, we obtain $x \in S$ with $x > L - \varepsilon = b$, contradicting the fact that $b$ is an upper bound. Hence $b \ge L$, so $L$ is the least upper bound, i.e. $L = \sup S$.
</Solution>
</Exercise>

<Exercise id="exr-nested-intervals" difficulty="Standard">
Let $I_n = [a_n, b_n]$ (with $a_n \le b_n$) be a sequence of closed intervals with $I_{n+1} \subset I_n$ for every $n$ and with $b_n - a_n \to 0$. Using <Ref to="ax-completeness" />, show that $\bigcap_{n \ge 1} I_n$ consists of exactly one point (the nested interval principle).

<Solution>
**Preliminaries.** The condition $I_{n+1} \subset I_n$ is equivalent to $a_n \le a_{n+1}$ and $b_{n+1} \le b_n$. Hence $(a_n)$ is increasing and $(b_n)$ is decreasing. Moreover $a_m \le b_n$ for all $m, n$: indeed, if $m \ge n$ then $a_m \le b_m \le b_n$, and if $m < n$ then $a_m \le a_n \le b_n$.

**Existence of the limit.** The sequence $(a_n)$ is increasing and, by the observation above, has $b_1$ as an upper bound. By <Ref to="ax-completeness" /> there is an $L$ with $a_n \to L$. Since $b_n = a_n + (b_n - a_n)$ and $b_n - a_n \to 0$ by hypothesis, for any $\varepsilon > 0$ and $n$ large enough we get $|b_n - L| \le |b_n - a_n| + |a_n - L| < \varepsilon$, so $b_n \to L$.

**$L$ lies in the intersection.** Fix $n$. Since $a_m \le b_n$ for every $m$, we have $L \le b_n$ (if $L > b_n$, then applying $\varepsilon = L - b_n > 0$ would give $a_m > L - \varepsilon = b_n$ for large $m$, a contradiction). Similarly $a_n \le L$ (if $a_n > L$, then $a_m \ge a_n > L$ for all $m \ge n$, so $|a_m - L| \ge a_n - L > 0$ persists, contradicting $a_m \to L$). Hence $L \in I_n$, and as $n$ was arbitrary, $L \in \bigcap_n I_n$.

**Uniqueness.** Suppose $L, L' \in \bigcap_n I_n$ with $L \ne L'$, and set $d = |L - L'| > 0$. As both lie in $I_n$, we have $d \le b_n - a_n$ for every $n$; but $b_n - a_n \to 0$ gives an $n$ with $b_n - a_n < d$, a contradiction. So the intersection is the single point $\{L\}$.
</Solution>
</Exercise>

<Exercise id="exr-recursive-sequence" difficulty="Standard">
Show that the sequence defined by $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$ (for $n \ge 1$) converges, and find its limit.

<Solution>
**Preliminaries.** The existence of $\sqrt{c}$ for a positive real $c$ is obtained by the same argument as in the proof of <Ref to="prop-sqrt-two" /> with $2$ replaced by $c$, so we may use it below.

**Boundedness.** We show $0 < a_n < 2$ for every $n$ by induction. For $n = 1$ we have $0 < 1 < 2$. Assuming $0 < a_n < 2$, we get $2 < 2 + a_n < 4$, so by monotonicity of the square root (for positive $u, v$, $u^2 < v^2 \iff u < v$) we get $\sqrt{2} < a_{n+1} < 2$, and in particular $0 < a_{n+1} < 2$.

**Monotonicity.** We have $a_{n+1}^2 - a_n^2 = (2 + a_n) - a_n^2 = -(a_n - 2)(a_n + 1)$. From $0 < a_n < 2$ we get $a_n - 2 < 0$ and $a_n + 1 > 0$, so this quantity is positive. Since $a_{n+1} > 0$ and $a_n > 0$, this gives $a_{n+1} > a_n$. Hence $(a_n)$ is increasing.

**Convergence.** Being increasing with upper bound $2$, the sequence has a limit $L$ by <Ref to="ax-completeness" />. From $a_n \ge a_1 = 1$ we get $L \ge 1 > 0$ (if $L < 1$, then for $\varepsilon = 1 - L > 0$ the inequality $|a_n - L| \ge a_1 - L = \varepsilon$ would hold for every $n$, contradicting convergence).

**Determining the limit.** Square the recursion to get $a_{n+1}^2 = 2 + a_n$ and let $n \to \infty$. The sequence $(a_{n+1})$ is $(a_n)$ with shifted indices, so it converges to the same $L$ (use the same $N$ for a given $\varepsilon$: if $n \ge N$ then $n + 1 \ge N$, so $|a_{n+1} - L| < \varepsilon$). By the limit laws for products and sums (see [Limits and continuity (the ε-δ argument)](/en/mathematics/calculus/limits-and-continuity)) we get $L^2 = 2 + L$, i.e. $(L - 2)(L + 1) = 0$. Since $L > 0$, we conclude $L = 2$.
</Solution>
</Exercise>

<Exercise id="exr-consecutive-differences" difficulty="Hard">
Prove the following two statements.

1. For $a_n = \sqrt{n}$ we have $|a_{n+1} - a_n| \to 0$, yet $(a_n)$ is not a Cauchy sequence. Thus "the differences of consecutive terms tend to $0$" is not a sufficient condition for being a Cauchy sequence.
2. If there are constants $C > 0$ and $0 < r < 1$ with $|a_{n+1} - a_n| \le C r^n$ for every $n$, then $(a_n)$ is a Cauchy sequence and therefore converges.

<Solution>
**(1)** Rationalising,
$$
\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \le \frac{1}{2\sqrt{n}} .
$$
Given $\varepsilon > 0$, use <Ref to="thm-archimedes" />(1) to take $N > \dfrac{1}{4\varepsilon^2}$. Then for $n \ge N$ we have $2\sqrt{n} > 2 \cdot \dfrac{1}{2\varepsilon} = \dfrac{1}{\varepsilon}$, so $|a_{n+1} - a_n| < \varepsilon$, proving $|a_{n+1} - a_n| \to 0$.

On the other hand $(\sqrt{n})$ is not bounded. Given any $M > 0$, <Ref to="thm-archimedes" />(1) supplies an $n$ with $n > M^2$, and then $\sqrt{n} > M$. By the contrapositive of <Ref to="prop-cauchy-basic" />(2), an unbounded sequence is not a Cauchy sequence.

**(2)** First we show $r^n \to 0$ for $0 < r < 1$. Since $\dfrac{1}{r} > 1$ we may write $\dfrac{1}{r} = 1 + h$ with $h > 0$. Bernoulli's inequality $(1 + h)^n \ge 1 + nh$ (by induction on $n$: equality for $n = 1$, and $(1+h)^{n+1} \ge (1 + nh)(1 + h) = 1 + (n+1)h + nh^2 \ge 1 + (n+1)h$) gives
$$
0 < r^n = \frac{1}{(1+h)^n} \le \frac{1}{1 + nh} < \frac{1}{nh} .
$$
Given $\varepsilon > 0$, taking $n > \dfrac{1}{h\varepsilon}$ by <Ref to="thm-archimedes" />(1) gives $r^n < \varepsilon$, so $r^n \to 0$.

Next let $m > n$. Applying the triangle inequality repeatedly and using the formula for a geometric sum,
$$
|a_m - a_n| \le \sum_{k=n}^{m-1} |a_{k+1} - a_k| \le \sum_{k=n}^{m-1} C r^k = C r^n \cdot \frac{1 - r^{m-n}}{1 - r} \le \frac{C r^n}{1 - r}
$$
(the last inequality because $0 < 1 - r^{m-n} < 1$). Given $\varepsilon > 0$, since $r^n \to 0$ we may take $N$ with $r^N < \dfrac{\varepsilon(1 - r)}{C}$. For $m, n \ge N$: if $m = n$ then $|a_m - a_n| = 0 < \varepsilon$; if $m \ne n$, letting $m$ denote the larger index, the estimate above gives
$$
|a_m - a_n| \le \frac{C r^n}{1 - r} \le \frac{C r^N}{1 - r} < \varepsilon
$$
(here $r^n \le r^N$ because $0 < r < 1$ and $n \ge N$). Hence $(a_n)$ is a Cauchy sequence and converges by <Ref to="thm-completeness" />.
</Solution>
</Exercise>

## References

- Teiji Takagi, *Kaiseki Gairon* (Introduction to Analysis), revised 3rd ed., Iwanami Shoten (in Japanese) — Chapter 1, "Basic concepts". A classic treatment that introduces the continuity of the reals by cuts and builds the basic convergence theorems from there.
- Mitsuo Sugiura, *Kaiseki Nyumon I* (Introduction to Analysis I), University of Tokyo Press, 1980 (in Japanese) — Chapter I. A detailed account of the axiomatic treatment of the reals and of the relations among suprema, the Archimedean property and the Bolzano–Weierstrass theorem.
- W. Rudin, *Principles of Mathematical Analysis*, 3rd ed., McGraw-Hill, 1976 — Chapter 1 (the real and complex number systems, characterisation by the least upper bound property), Chapter 3 (sequences and series, Cauchy sequences and completeness).
- Kazuo Matsuzaka, *Shugo, Iso Nyumon* (Introduction to Sets and Topology), Iwanami Shoten, 1968 (in Japanese) — the chapter constructing the reals from the rationals. It covers the construction by cuts and the uniqueness of the complete ordered field up to isomorphism.
- R. Dedekind, *Stetigkeit und irrationale Zahlen* (On Numbers: Continuity and Irrational Numbers), Japanese translation by Isaburo Kono, Iwanami Bunko (in Japanese) — a translation of the original 1872 memoir. The way the question "what is the continuity of a line?" is posed is itself worth reading.

## Appendix: equivalent formulations of completeness

**Equivalent formulations.** For an ordered field $K$ the following conditions are all equivalent. Whichever one is chosen as the axiom, the same field of real numbers results.

| Condition | Statement |
|---|---|
| Monotone convergence principle | Every increasing sequence bounded above converges (<Ref to="ax-completeness" /> of this article) |
| Least upper bound property | Every non-empty subset bounded above has a supremum (<Ref to="thm-supremum" />) |
| Dedekind cuts | If $K$ is split into two non-empty parts $A$, $B$ such that every element of $A$ is at most every element of $B$, then exactly one of the two — a greatest element of $A$ or a least element of $B$ — always exists |
| Bolzano–Weierstrass | Every bounded sequence has a convergent subsequence (<Ref to="thm-bolzano-weierstrass" />) |
| Nested intervals + Archimedean property | A nested sequence of closed intervals has a common point, and the Archimedean property holds |
| Cauchy completeness + Archimedean property | Every Cauchy sequence converges, and the Archimedean property holds |

**Why the Archimedean property is required separately.** In the last two rows the Archimedean property is added as an extra condition. Dropping it destroys the equivalence. Indeed, there are known ordered fields that are non-Archimedean and yet in which every Cauchy sequence converges (a field of formal Laurent series with a suitable order is one example), so the Archimedean property cannot be derived from Cauchy completeness alone. By contrast, the monotone convergence principle and the least upper bound property adopted in this article derive the Archimedean property on their own, as we saw in <Ref to="thm-archimedes" />. This asymmetry is worth remembering when gauging the "strength" of a form of completeness.

**Existence and uniqueness.** So far we have argued on the assumption that a complete ordered field $\mathbb{R}$ exists. Its existence is guaranteed by a construction from the rational field (Dedekind cuts, or completion by equivalence classes of Cauchy sequences). Furthermore, a complete ordered field is known to be unique up to isomorphism of ordered fields. That is why we may speak of *the* real numbers. For the details of the constructions, see the references above, in particular Matsuzaka's *Shugo, Iso Nyumon* and Dedekind's original memoir.