5.2 Cauchy sequences and another model of the reals

The previous note defined convergence by comparing sequence terms with a previously known limit $L$. This note asks a harder question:

What if we want to recognize convergence before we already know the limit as an existing real number?

That question leads to the notion of a Cauchy sequence and to a second construction of the real numbers.

Why we need an internal test for convergence

Consider rational approximations obtained by adding successively smaller fractions:

$$1,\qquad 1+\frac12,\qquad 1+\frac12+\frac1{3\cdot 2},\qquad 1+\frac12+\frac1{3\cdot 2}+\frac1{4\cdot 3\cdot 2\cdot 1},\ \ldots$$

Precisely, the term with index $n\ge0$ is $x_n=\sum_{k=1}^{n+1}1/k!$. Every term can be computed using rational arithmetic. Naming a real limit would require an additional existence argument; the list itself does not supply one.

So instead of asking whether the sequence is close to some external point $L$, we ask whether the terms are getting close to each other.

Throughout the rational construction below, $x_n\in\mathbb Q$, $\varepsilon\in\mathbb Q_{\gt 0}$, and $N\in\mathbb N=\{0,1,2,\ldots\}$. We later use the already constructed Dedekind real field explicitly to verify completeness.

The definition of Cauchy sequence

Definition

Cauchy sequence

A sequence $(x_n)$ is Cauchy if for every $\varepsilon\gt 0$, there exists $N$ such that

$$|x_n-x_m|\lt \varepsilon$$

for all $n,m\gt N$.

The same threshold must work for every pair of later indices, even when they are far apart. First fix $\varepsilon$, then choose $N$, then consider arbitrary $n,m>N$. Making only successive differences $|x_{n+1}-x_n|$ small does not by itself control the distance across a long tail.

This definition has the same quantifier shape as the limit definition, but the comparison target has changed:

• for an ordinary limit, you compare $x_n$ with a fixed number $L$;
• for a Cauchy condition, you compare late terms of the sequence with one another.

So a Cauchy sequence is one whose tail fits into narrower and narrower bands.

Common mistake

Cauchy does not mean monotone

A Cauchy sequence does not have to move only upward or only downward. The definition says nothing about monotonicity. It only says that late terms become uniformly close to one another.

Worked example

Control a tail without naming its limit

For the rational partial sums above and $n>m\ge0$, the bound $k!\ge2^{k-1}$ gives

$$0<x_n-x_m =\sum_{k=m+2}^{n+1}\frac1{k!} \le\sum_{j=m+1}^{n}2^{-j} <2^{-m}\le\frac1{m+1}.$$

The strict bound uses the finite geometric sum, while $2^m\ge m+1$ follows by induction. Given rational $\varepsilon>0$, choose a natural $N>1/\varepsilon$. If $n,m>N$, exchange them if needed so that $n\ge m$. Equal indices give error zero; otherwise the displayed bound is below $\varepsilon$. Thus the sequence is Cauchy using only rational estimates, before any limit is named.

The construction now has three jobs: decide when two such approximations represent the same number, make arithmetic independent of that choice, and prove that the resulting number system is complete.

Why convergent sequences are automatically Cauchy

The key proposition is the following.

Theorem

If a sequence converges, then it is Cauchy

Suppose a rational sequence $(x_n)$ has a rational limit $L$. Then $(x_n)$ is a Cauchy sequence.

The proof idea is short and very important.

Proof using the triangle inequality

Assume $x_n\to L$, and let $\varepsilon\gt 0$ be given.

Because $x_n\to L$, there exists $N$ such that for every $n\gt N$,

$$|x_n-L|\lt \frac{\varepsilon}{2}.$$

Then whenever $n,m\gt N$, the triangle inequality gives

$$|x_n-x_m| = |(x_n-L)+(L-x_m)| \le |x_n-L|+|x_m-L| \lt \frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.$$

So $(x_n)$ is Cauchy.

This theorem says that genuine convergence always forces the sequence tail to compress.

Equivalent Cauchy sequences

If Cauchy sequences are going to represent real numbers, then different sequences that “head toward the same place” should count as the same real.

Definition

Equivalent Cauchy sequences

Two Cauchy sequences $(x_n)$ and $(y_n)$ are equivalent if for every $\varepsilon\gt 0$, there exists $N$ such that

$$|x_n-y_m|\lt \varepsilon$$

for all $n,m\gt N$.

This condition says that the two tails eventually lie arbitrarily close to one another. Intuitively, they are describing the same limiting point on the line.

The equivalence relation and its same-index form

The rational triangle inequality follows by adding $-|a|\le a\le|a|$ and $-|b|\le b\le|b|$: their sum lies between $-(|a|+|b|)$ and $|a|+|b|$, so $|a+b|\le|a|+|b|$.

Reflexivity of the relation above is the Cauchy condition; symmetry follows by swapping indices. For transitivity, suppose $x\sim y$ and $y\sim z$. Choose a common threshold for tolerance $\varepsilon/2$, then fix any $k$ beyond it. For arbitrary $n,m$ beyond it,

$$|x_n-z_m|\le|x_n-y_k|+|y_k-z_m|\lt \varepsilon.$$

Thus the relation is an equivalence relation. It is equivalent to $x_n-y_n\to0$, with rational tolerances. One direction sets $m=n$. Conversely, use

$$|x_n-y_m|\le|x_n-y_n|+|y_n-y_m|$$

and allocate half the tolerance to each term. The second term uses the Cauchy property of $y$. Changing finitely many terms preserves the class, since the threshold can be enlarged beyond all changed indices. Distinct rational constants give distinct classes: tolerance $|q-r|/2$ rules out equivalence when $q\ne r$.

The alternative construction of $R$

Now turn the idea into a definition.

Definition

Reals as equivalence classes of rational Cauchy sequences

Let $F$ (the Cauchy model of $R$) be the set of equivalence classes of Cauchy sequences of rational numbers. These equivalence classes form another model of the real numbers.

This is a big shift in viewpoint:

• in the Dedekind-cut model, a real number is a left/right split of $Q$;
• in the Cauchy-sequence model, a real number is a whole family of rational sequences that become indistinguishable in the limit.

Neither model is “more real” than the other. They are two rigorous ways to build the same number system.

How rationals sit inside the model

It remains to say how rational numbers sit inside this construction. The answer is natural: a rational $q$ is represented by the constant Cauchy sequence

$$(q,q,q,q,\ldots).$$

Worked example

Representatives of $1/2$

The real number $1/2$ can be represented by the constant sequence

$$\left(\frac12,\frac12,\frac12,\ldots\right).$$

It can also be represented by other rational Cauchy sequences that converge to the same point, for example

$$\left(\frac12+\frac11,\frac12+\frac12,\frac12+\frac13,\frac12+\frac14,\ldots\right).$$

The second sequence is not constant, but its terms get arbitrarily close to $1/2$, so it belongs to the same equivalence class.

The real number is therefore not any one representative sequence by itself. It is the full equivalence class.

Common mistake

A real number is an equivalence class, not a favourite representative

Once this model is adopted, changing from one representative Cauchy sequence to another equivalent one does not change the real number. The representative is a description, not the object itself.

Boundedness and rational arithmetic on classes

Every sequence in this construction has rational terms indexed by $\mathbb N=\{0,1,2,\ldots\}$. Until we explicitly introduce the Dedekind real field below, every tolerance is in $\mathbb Q_{\gt 0}$ and every threshold is in $\mathbb N$.

Every Cauchy sequence is bounded

Choose $N_0$ so that pairwise differences beyond it are less than $1$, and set $k=N_0+1$. For every $n\gt N_0$, $|x_n|\lt |x_k|+1$. The finitely many earlier terms $x_0,\ldots,x_{N_0}$ have a rational absolute-value bound. Taking the maximum of these bounds and $1$ produces a rational global bound $M\ge1$. The anchor $k$ must be strictly beyond the Cauchy threshold.

Worked example

Bounded, unbounded, and nonmonotone sequences

The bounded sequence $x_n=(-1)^n$ is not Cauchy. Use tolerance $1$. Given any threshold $N$, choose an even $m\gt N$ and an odd $n\gt N$; then $|x_m-x_n|=2\gt1$. Boundedness alone therefore does not force the Cauchy condition.

The sequence $x_n=n$ is unbounded and also fails the Cauchy condition: with tolerance $1$, the adjacent terms $x_{N+1}$ and $x_{N+2}$ differ by exactly $1$. In contrast, $(-1)^n/(n+1)$ is not monotone, but it converges to zero and is Cauchy. These examples separate boundedness, monotonicity, and the actual tail condition in the definition.

Closure and independence of representatives

Write $F$ for the quotient set. Define

$$[x]+[y]=[(x_n+y_n)],\qquad [x][y]=[(x_ny_n)],\qquad -[x]=[(-x_n)].$$

We must check both closure and independence of representatives. For addition, a difference of sums is bounded by the two original differences; make each less than $\varepsilon/2$ on a common tail. For multiplication, first choose a rational common bound $M\ge1$ for both sequences. Then

$$|x_ny_n-x_my_m|\le |x_n||y_n-y_m|+|y_m||x_n-x_m| \le M|y_n-y_m|+M|x_n-x_m|\lt \varepsilon$$

once each original difference is less than $\varepsilon/(2M)$. Thus the product is Cauchy. Negation preserves absolute differences directly.

If $x\sim x'$ and $y\sim y'$, use the same addition argument with the cross-index differences. For products, choose one rational $M\ge1$ bounding all four sequences. On a common tail, equivalence makes each difference below $\varepsilon/(2M)$, giving

$$|x_ny_n-x'_my'_m|\le M|y_n-y'_m|+M|x_n-x'_m|\lt \varepsilon.$$

Hence the operations are well-defined. Associativity, commutativity, and distributivity hold term by term by rational arithmetic. Constant sequences $0$ and $1$ give the identities; $(-x_n)$ gives the additive inverse.

Worked example

Nonzero terms can still represent the zero class

Let $x_n=1/(n+1)$. Every term is nonzero, but the sequence represents $[0]$: given rational $\varepsilon\gt0$, choose $N$ with $1/(N+1)\lt\varepsilon$; then $|x_n|\lt\varepsilon$ for every $n\gt N$. Its termwise reciprocal is $u_n=n+1$, which is not Cauchy, since for tolerance $1$ the adjacent late terms always differ by exactly $1$. Thus nonzero terms alone do not justify taking reciprocals; a nonzero class requires an eventual uniform lower bound away from zero.

Multiplicative inverses require a uniform lower bound

Suppose $[x]\ne[0]$. There must be rational $c\gt 0$ and natural $N$ such that $|x_n|\ge c$ for every $n\gt N$. Otherwise, given rational $\varepsilon\gt 0$, first choose a Cauchy threshold $N_0$ for $\varepsilon/2$. The contrary assumption gives $k\gt N_0$ with $|x_k|\lt \varepsilon/2$. For every $n\gt N_0$, the triangle inequality then gives $|x_n|\lt \varepsilon$. This says $x\sim0$, a contradiction.

Define $u_n=1/x_n$ for $n\gt N$, and set the earlier terms to $0$. The reciprocal sequence is Cauchy: beyond this threshold,

$$|u_n-u_m|=\frac{|x_n-x_m|}{|x_nx_m|}\le\frac{|x_n-x_m|}{c^2}\lt \varepsilon$$

when the original difference is below the rational tolerance $c^2\varepsilon$.

To check independence of representatives, let $x'\sim x$ and choose an eventual rational lower bound $d\gt 0$ for $|x'_m|$. On a common tail,

$$\left|\frac1{x_n}-\frac1{x'_m}\right|\le\frac{|x_n-x'_m|}{cd}\lt \varepsilon$$

by equivalence with tolerance $cd\varepsilon$. Finite initial choices do not matter either. Finally, $x_nu_n=1$ eventually, so $[x][u]=[1]$. This proves existence of a well-defined inverse. Uniqueness follows from associativity: if $ab=ac=1$, then $b=b(ac)=(ba)c=c$.

Order must be independent of representatives

Plain eventual pointwise comparison fails this requirement. The equivalent sequences $0$ and $1/(n+1)$ have different pointwise comparisons with zero. Instead define

$$[x]\le_F[y]\quad\Longleftrightarrow\quad \forall\varepsilon\in\mathbb Q_{\gt 0}\ \exists N\in\mathbb N\ \forall n\gt N,\quad x_n\le y_n+\varepsilon.$$

If representatives change to $x'$ and $y'$, allocate $\varepsilon/3$ to $|x'_n-x_n|$, the original order tolerance, and $|y_n-y'_n|$. On a common tail this yields $x'_n\le y'_n+\varepsilon$. Reversing the replacements proves independence. We will verify all order properties by showing that this relation agrees exactly with the order of the previously constructed real field.

Theorem

Endpoint theorem: the Cauchy model is the Dedekind real field

After the verification below, the map

$$\Phi:F\longrightarrow R_D,\qquad \Phi([x])=\lim_{n\to\infty}x_n$$

is an order-preserving field isomorphism that fixes every embedded rational. The proof explicitly uses the already constructed Dedekind complete ordered field $R_D$, including its least-upper-bound property, rational density, and Archimedean property. It is therefore an identification with the earlier Dedekind model, not an independent construction of completeness from the Cauchy definitions alone.

Verification using the Dedekind real field

Let $R_D$ be the complete ordered field constructed in Chapter 4. The quotient definitions above use only rational data; the following verification of completeness uses the earlier Dedekind construction. It is not an independent proof of completeness from scratch. We use the least-upper-bound property and rational density of $R_D$. Positive rational tolerances suffice for real errors, because every positive real tolerance has a smaller positive rational number.

A rational Cauchy sequence converges in $R_D$

Regard a bounded rational Cauchy sequence as a sequence in $R_D$. Define

$$\ell_N=\inf_{n\gt N}x_n,\qquad u_N=\sup_{n\gt N}x_n,\qquad L=\sup_{N\in\mathbb N}\ell_N.$$

Every tail is nonempty and bounded. Thus its infimum and supremum exist, and the increasing, bounded sequence of lower bounds has a supremum $L$. For each $N$, $\ell_N\le L\le u_N$: every $\ell_K$ is at most $u_N$, as can be seen by choosing a term whose index exceeds both $K$ and $N$.

Given a positive real tolerance $\eta$, choose rational $0\lt \rho\lt \eta$ and a Cauchy threshold $N$ for $\rho$. Fix $m\gt N$. All tail terms satisfy $x_n\lt x_m+\rho$, so $u_N\le x_m+\rho$. Thus $u_N-\rho$ is a lower bound of the tail, so $u_N-\rho\le\ell_N$, or $u_N\le\ell_N+\rho$. Both $L$ and each later $x_n$ belong to $[\ell_N,u_N]$, hence $|x_n-L|\le\rho\lt \eta$. This proves convergence without assuming a Cauchy convergence theorem.

Limits in $R_D$ are unique: if distinct $L,M$ were both limits, use tolerance $|L-M|/3$ for each. The triangle inequality would give $|L-M|\lt 2|L-M|/3$, a contradiction.

A bijection between the two models

Define

$$\Phi:F\longrightarrow R_D,\qquad \Phi([x])=\lim_{n\to\infty}x_n.$$

If $x\sim y$ with limits $L,M$, split any positive real tolerance into three parts: control $|L-x_n|$, $|x_n-y_n|$, and $|y_n-M|$ on a common tail. Use a smaller positive rational tolerance for the middle term. Thus $|L-M|$ is smaller than every positive tolerance, so $L=M$; the map is well-defined. Conversely, if both limits equal $L$, then $|x_n-y_m|\le|x_n-L|+|y_m-L|\lt \varepsilon$ on a common tail. Hence $x\sim y$, proving injectivity.

For $r\in R_D$, rational density gives $q_n\in\mathbb Q$ with $|q_n-r|\lt 1/(n+1)$. A definite choice uses the integer-fence property from Chapter 4: take the unique integer $k_n$ with $k_n\le(n+1)r\lt k_n+1$ and put $q_n=k_n/(n+1)$. The Archimedean property gives $q_n\to r$; the triangle inequality makes $(q_n)$ Cauchy. Therefore $\Phi([q])=r$, proving surjectivity. Constants show that $\Phi$ fixes the embedded rationals.

Preservation of arithmetic and order

Suppose $x_n\to L$ and $y_n\to M$. The estimate

$$|(x_n+y_n)-(L+M)|\le|x_n-L|+|y_n-M|$$

with half the desired tolerance for each term proves preservation of addition. For multiplication, take a rational bound $B\ge1$ for $|x_n|$ and set $K=\max(B,|M|,1)$ in $R_D$. Then

$$|x_ny_n-LM|\le |x_n||y_n-M|+|M||x_n-L| \le K\bigl(|y_n-M|+|x_n-L|\bigr).$$

For a positive real tolerance $\eta$, make each difference less than $\eta/(2K)$ on a common tail. This proves product convergence directly, without appealing to unproved sequence limit laws. Therefore $\Phi$ preserves multiplication, and it preserves $0$ and $1$ by constants.

If $L\le M$, for any rational $\varepsilon\gt 0$ choose a common tail with both limit errors below $\varepsilon/2$. Then $x_n\lt y_n+\varepsilon$, so $[x]\le_F[y]$. Conversely, if $[x]\le_F[y]$ but $L\gt M$, choose rational $0\lt \rho\lt (L-M)/3$. Once both limit errors are less than $\rho$, we get $x_n-y_n\gt L-M-2\rho\gt \rho$. This contradicts the order definition with tolerance $\rho$. Hence

$$[x]\le_F[y]\quad\Longleftrightarrow\quad\Phi([x])\le\Phi([y]).$$

Transfer of the least-upper-bound property

The bijection preserves arithmetic and order, so $F$ is an ordered field. Let $S\subseteq F$ be nonempty and bounded above by $b$. Its image is nonempty and bounded above by $\Phi(b)$ in $R_D$. Set $t=\sup\Phi(S)$ and $s=\Phi^{-1}(t)$. Order preservation shows that $s$ bounds $S$. If $v$ is any upper bound of $S$, then $\Phi(v)$ bounds $\Phi(S)$, so $t\le\Phi(v)$ and $s\le v$. Therefore $s=\sup S$.

This completes the verification that the rational Cauchy quotient is a complete ordered field. Its completeness has been established by identification with the Dedekind model, while its definitions, representative checks, and inverse estimates were rational throughout.

Quick checks

Checkpoint

What is the difference between the definitions of ‘convergent’ and ‘Cauchy’?

Focus on what each definition compares $x_n$ with.

Solution · Answer

A convergent sequence compares each late term $x_n$ with a fixed limit $L$, while a Cauchy sequence compares late terms $x_n$ and $x_m$ directly with one another.

Checkpoint

How does the rational number q appear inside the Cauchy-sequence model of R?

Think of the simplest possible Cauchy sequence.

Solution · Answer

It appears as the equivalence class of the constant sequence $(q,q,q,q,\ldots)$.

Checkpoint

Why is boundedness useful when proving that products of Cauchy sequences are Cauchy?

Look at the estimate for $|x_my_m-x_ny_n|$.

Solution · Answer

Boundedness lets us replace $|x_m|$ and $|y_n|$ by a common constant $M$, so the product difference can be controlled by the small Cauchy differences $|x_m-x_n|$ and $|y_m-y_n|$.

Exercises

Checkpoint

Show that every constant rational sequence is Cauchy.

Use the fact that all pairwise differences are zero.

Solution · Guided solution

Let $x_n=q$ for all $n$, where $q\in Q$. Then for every $n,m$,

$$|x_n-x_m|=|q-q|=0.$$

So for any $\varepsilon\gt 0$, every $N$ works. Hence every constant rational sequence is Cauchy.

Checkpoint

Let $q$ be rational and $x_n=q+1/(n+1)$. Prove directly that (x_n) is equivalent to the constant sequence $q$.

Write a bound valid for every pair of indices beyond one threshold.

Solution · Model solution

For every $n,m$, $|x_n-q|=1/(n+1)$. Given rational $\varepsilon>0$, choose $N>1/\varepsilon$. Every $n,m>N$ then satisfies $|x_n-q|<\varepsilon$. The sequence $(x_n)$ is Cauchy because it converges to the rational $q$, and the cross-index condition proves equivalence. The index $m$ causes no additional error because its representative is constant.

Checkpoint

The sequences $x_n=0$ and $y_n=1/(n+1)$ are equivalent. Explain why eventual termwise order would fail to define an order on their classes.

Compare both $[y]\le[0]$ and the proposed condition $y_n\le0$ eventually.

Solution · Model solution

The classes satisfy $[y]=[0]$, so reflexivity requires $[y]\le[0]$. Yet $y_n>0$ for every $n$, whereas the equivalent representative $x_n=0$ satisfies $x_n\le0$ for every $n$. Eventual pointwise comparison therefore changes when the representative changes. The tolerance-based order in this note fixes the problem by allowing an arbitrarily small positive error on a sufficiently late tail.

Prerequisites and continuation

Read this after 5.1 Sequences and epsilon-N limits and 4.3 Completeness and gaps in Q. Then continue to 5.3 Delta-epsilon limits, limit laws, and continuity.

The completeness verification also uses 4.5 Dedekind cuts and embedding of Q.