4.3  Sequences and Limits in an Ordered Field

In $\mathbb{Q}$ the fractions $1/2,2/3,3/4,\dots$ “tend to $1$”. Sections 4.1 and 4.2 supplied the tools to say what that means without leaving $\mathbb{Q}$: an absolute value to measure distance, and the Archimedean property to make $1/(n+1)$ small. This section defines the limit of a sequence in an arbitrary ordered field, with tolerances taken from that field. It then proves that a sequence has at most one limit, and that a convergent sequence is bounded. Chapter 5 applies these results with $F=\mathbb{Q}$, at a stage when no other number system is available.

Throughout, $F$ is an ordered field (Definition 2.3.2), and $F_{>0}=\{x\in F:x>0\}$ is its set of positive elements, as in Section 4.1; for $F=\mathbb{Q}$ this is ${\mathbb{Q}}_{>0}$. As agreed after Proposition 4.2.1, we write $n$ for $n\cdot 1_F$ where no confusion arises. So $1/(n+1)$ in $F$ means $1/((n+1)\cdot 1_F)$, which exists and is positive, since $(n+1)\cdot 1_F\ge 1>0$ by Proposition 4.2.1 (b) and reciprocals of positive elements are positive (Proposition 2.5.2 (a)).

4.3.1. Sequences

A list $1,1/2,1/3,\dots$ is a rule that assigns a term to each position, and a rule of that kind already has a name: a function on $\mathbb{N}$.

Examples.

• In $\mathbb{Q}$: $a_n:=1/(n+1)$, with terms $1,1/2,1/3,\dots$.
• In any $F$: $a_n:=n$, that is, $n\cdot 1_F$, with terms $0,1,2,\dots$; and $a_n:=(-1{)}^n$, with the powers of Definition 4.2.1, with terms $1,-1,1,\dots$. The second sequence has infinitely many terms but only two values: its set of values is $\{1,-1\}$. A sequence is ordered and may repeat; its set of values forgets both.
• The degenerate case: a constant sequence $a_n:=c$ for a fixed $c\in F$. It matters because every definition below must give the expected answer on it.
• A sequence may be defined by recursion (Sets, Relations, and Cardinality, Theorem 2.3.1, read on $\mathbb{N}=\omega$ as Theorem 1.5.1 through Corollary 1.7.3 (b)). The Newton sequence of Example (A Rational Sequence Aimed at the Gap), $x_0=2$ and $x_{n+1}=(x_n^2+2)/(2x_n)$, is a sequence in $\mathbb{Q}$; Section 4.6 studies it.

Non-example by minimal change. The formula $n\mapsto 1/n$ defines no sequence: at $n=0$ it asks for $1/0$, and $0$ has no inverse. Everything else about the first example survives the change, and only “a function on all of $\mathbb{N}$” fails. This is why the book’s sequences start at $n=0$ and why we write $1/(n+1)$.

“Eventually” absorbs any finite beginning. In $\mathbb{Q}$, “$1/(n+1)<1/10$” holds eventually, from $N_0=10$ on, although it fails for $n\le 9$; “$(-1{)}^n=1$” does not hold eventually, since it fails at every odd $n$. Two eventual properties hold together eventually: if $P(n)$ holds for $n\ge N_0$ and $Q(n)$ for $n\ge N_1$, then both hold for $n\ge max\{N_0,N_1\}$, the larger of the two numbers, which exists by trichotomy in $\mathbb{N}$ (Theorem 1.6.1 (a)). The same goes for any finite number of properties, taking the larger number one comparison at a time. For infinitely many properties it fails: “$1/(n+1)<1/(k+1)$” holds eventually for each $k\in \mathbb{N}$, from $N_0=k+1$ on, but no single $N_0$ serves every $k$, since $k=N_0$ fails at $n=N_0$.

4.3.2. The limit of a sequence

We want a precise sense in which $1/2,2/3,3/4,\dots$ “tends to $1$” in $\mathbb{Q}$. The first idea is “the terms get closer and closer to $1$”. Here is why it fails. With $a_n:=(n+1)/(n+2)$, $$1-a_n=\frac{1}{n+2},2-a_n=\frac{n+3}{n+2}=1+\frac{1}{n+2}.$$ Both distances shrink at every step, so the terms get “closer and closer” to $2$ as well. What singles out $1$ is that the distance to $1$ falls below any tolerance we name, and stays below it, while the distance to $2$ never falls below $1$. The tolerances must come from $F$ itself, since in $\mathbb{Q}$ there is nothing else to measure with.

Every tolerance in $F$ is eventually met.

In words: the first clause hands us an arbitrary tolerance $\epsilon$, a positive element of $F$. The second says that some index $N$ can be found, and it may depend on $\epsilon$. The third says that from $N$ on, every term lies within $\epsilon$ of $a$: the inequality holds eventually. The absolute value is that of Definition 4.1.1, and by Proposition 4.1.3 (d) the condition on each term reads $$|a_n-a|<\epsilon ⟺a-\epsilon <a_n<a+\epsilon .$$

The domain of $\epsilon$ is part of the definition. For $F=\mathbb{Q}$, $\epsilon$ is a rational number, and “$a_n\to a$ in $\mathbb{Q}$” mentions nothing outside $\mathbb{Q}$. The formula is the one translated in Logic and Proof, Example (Five definitions, translated) (a), where it was read in the school real numbers. Here it is read in $F$, and every fact used about it will be proved from the ordered-field axioms.

Well-definedness. Nothing needs checking for “converges to $a$”. We say “a limit” for now, because a sequence might have two; Theorem 4.3.4 below rules that out.

Negating the definition, quantifier by quantifier (Logic and Proof, Proposition 6.2.3 for the bounded quantifiers, Logic and Proof, Proposition 6.2.2 for the implication), and using that $\mathrm{\neg }(x<\epsilon )$ means $x\ge \epsilon$ in the totally ordered $F$ (Proposition 4.1.2 (a)): $a_n\nrightarrow a$ means $$\mathrm{\exists }\epsilon \in F_{>0}\mathrm{\forall }N\in \mathbb{N}\mathrm{\exists }n\in \mathbb{N}(n\ge N\wedge |a_n-a|\ge \epsilon ).$$ One bad tolerance, and beyond every threshold a term that misses it. A sequence is divergent when this holds for every $a\in F$.

Non-example by minimal change. Keep the sequence $a_n=(n+1)/(n+2)$ and change the candidate limit from $1$ to $2$. The distance $|a_n-2|=1+1/(n+2)$ still shrinks at every step, and still the definition fails, at its first clause: for $\epsilon :=1$, every term has $|a_n-2|>1$, so no $N$ exists. This is the negation displayed above, with $\epsilon =1$ and $n=N$ for every $N$.

Why this definition.

• Strict or not. Replacing $|a_n-a|<\epsilon$ by $|a_n-a|\le \epsilon$ gives the same notion: given $\epsilon$, apply the weak form to $\epsilon /2$, and $\epsilon /2<\epsilon$ (Proposition 4.1.2 (b)).
• Every $n\ge N$, not some. With “some $n\ge N$” the sequence $(-1{)}^n$ would “converge” both to $1$ and to $-1$: beyond any $N$ there are terms equal to each.
• The order of the quantifiers. With $\mathrm{\exists }N$ before $\mathrm{\forall }\epsilon$, the condition says that $a_n=a$ for every $n\ge N$: for each such $n$, $|a_n-a|$ lies below every element of $F_{>0}$, so it is not itself positive, and it is $0$ by Proposition 4.1.3 (a). (Exercise A1 (b) asks for the details.) Then $1/(n+1)$ would have no limit in $\mathbb{Q}$.
• Tolerances from $F$. In a field with more elements than $\mathbb{Q}$, “every rational tolerance is eventually met” can be strictly weaker than “every tolerance in $F$ is eventually met”. The remark after Proposition 4.3.1 shows this happen.

Two kinds of sequence get their own names, because they recur throughout the rest of the chapter.

Examples. In $\mathbb{Q}$, $(1/(n+1))$ is null and bounded by $1$; in every $F$, $((-1{)}^n)$ is bounded by $1$ and not null (Example (First Limits) (d)); a constant sequence $c$ is bounded by $\left|c\right|+1$, which is positive even when $c=0$. Non-example. In $\mathbb{Q}$ the sequence $(n)$ is not bounded: for any $M\in {\mathbb{Q}}_{>0}$, Archimedes (Theorem 2.5.3) gives $n\in \mathbb{N}$ with $M<n=\left|n\right|$. The bound is required to be positive only for convenience, so that we may divide by it; any $M\ge 0$ that works can be replaced by $M+1$.

4.3.3. Archimedes and the sequence 1/(n + 1)

In $\mathbb{Q}$, $1/(n+1)\to 0$ came straight from Archimedes. In a general ordered field, the two statements turn out to be the same.

So in $\mathbb{Q}(t)$, the field of Example (A Non-Archimedean Ordered Field), which is not Archimedean, $1/(n+1)$ does not tend to $0$: the tolerance $1/t$ is never met. In $\mathbb{Q}$, $1/(n+1)\to 0$ is Example (First Limits) (b), which rests on the Archimedean property; every result proved without the Archimedean property holds in $\mathbb{Q}(t)$ as well.

4.3.4. Changing finitely many terms

Whether a sequence converges, and to what, depends only on its terms from some point on. Two forms of this are used constantly: altering finitely many terms, and dropping the first $k$.

The shift clause is about convergence only; it is used to start a sequence later, for instance to read a formula such as $1/n$, which needs $n\ge 1$, as the shift by $1$ of a sequence indexed from $0$. Section 4.5 adds the Cauchy property to clause (a).

One more reformulation moves every limit to $0$.

4.3.5. A sequence has at most one limit

If “the terms are eventually within every tolerance of $a$” is to name the number the sequence tends to, it must single out one $a$. It does.

Try it first (ten minutes): suppose $a_n\to a$ and $a_n\to b$ with $a\ne b$. Which $\epsilon$ makes the two tails collide?

Since a convergent sequence has exactly one limit, we may speak of the limit and write it $lim{a}_n$, or $\underset{n\to \mathrm{\infty }}{lim}{a}_n$. So $lim{a}_n=a$ means the same as $a_n\to a$. In $\mathbb{Q}$, Example (First Limits) gives $lim1/(n+1)=0$ and $lim(n+1)/(n+2)=1$, while $lim(-1{)}^n$ names nothing.

4.3.6. Convergent sequences are bounded

A convergent sequence cannot run off: from some index on, its terms are trapped near the limit, and only finitely many terms come before that index.

The bound $M$ depends on the whole sequence, through $a$, $N$ and the first $N$ terms, but on no tolerance: it is one element of $F$, fixed once. Section 4.4 uses exactly that, in the product of two limits. The converse fails: $((-1{)}^n)$ is bounded by $1$ and divergent (Example (First Limits) (d)).

4.3.7. Exercises

A. Check your understanding

B. Practice

C. Going deeper