Equivalent Forms of Completeness: Examples and Counterexamples

Let $\mathbb{R}(t)$ be the field of fractions of the polynomial ring $\mathbb{R}[t]$, and let

$P \;=\; \{\, f \in \mathbb{R}(t) : f \ne 0 \text{ and } f(x) > 0 \text{ for all sufficiently large real } x \,\}.$

Not every ordered field is Archimedean proves that $(\mathbb{R}(t), P)$ is an ordered field and that it is not Archimedean. This example works the order out in usable form. Three things are established below:

1. A computation rule. For $f = p/q$ with $p, q \in \mathbb{R}[t]$ nonzero, $f \in P$ exactly when $\operatorname{lc}(p)\operatorname{lc}(q) > 0$, where $\operatorname{lc}$ is the leading coefficient. So comparing two rational functions is comparing one product of two real numbers.
2. That the rule is independent of the representative $p/q$ chosen, which is what makes it a definition of a function on $\mathbb{R}(t)$ and not merely on pairs.
3. The two elements that make the field interesting: $t$, which exceeds every canonical natural, and $1/t$, which is positive and lies below every positive rational. An element of the second kind is called an infinitesimal, and its existence is exactly the failure of the Archimedean property (Archimedean ordered field).

• Why the eventual sign, and not the sign at a point. Evaluating at a fixed real $x_0$ is not even a function on all of $\mathbb{R}(t)$, since a rational function may have a pole at $x_0$; and even where evaluation is defined, its sign cannot give a positive cone on the field, because the nonzero rational function $t-x_0$ evaluates to $0$, so trichotomy fails. The behaviour at $+\infty$ is one representative-independent, multiplicative choice, and step 1.1 is exactly that statement.

• What this field is and is not good for. It is the library's cheapest witness that an ordered field need not be Archimedean, and In $\mathbb{R}(t)$ the rationals are not dense: no rational lies strictly between $0$ and $1/t$ uses the infinitesimal $1/t$ found above to show $\mathbb{Q}$ need not be dense. It is not a witness for the completeness failures of FALSE: the nested interval property alone implies the least-upper-bound property or FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property: nothing in this library proves that $\mathbb{R}(t)$ is Cauchy complete or that it has any nested interval property, and in fact it is neither. Those two need the larger field $\mathbb{R}((t^{-1}))$, which is why that field was built ($\mathbb{R}((t^{-1}))$, the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property).

• The order is the one induced from $\mathbb{R}((t^{-1}))$ in spirit but not by any embedding proved here. Both fields order an element by its behaviour at infinity, and in both the comparison looks at a single coefficient. This library constructs no embedding of one into the other and never uses one.

Let $u := \sqrt 2 \in \mathbb{R}$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$) and

$F \;:=\; \mathbb{Q}(\sqrt 2) \;=\; \{\, a + bu : a, b \in \mathbb{Q} \,\} \;\subseteq\; \mathbb{R}.$

Then $F$ is a field, every element of it is $a + bu$ for exactly one pair $(a,b)$ of rationals, and the conjugation $\sigma(a + bu) := a - bu$ is a field automorphism of $F$.

$F$ carries exactly two positive cones (Ordered field):

$P_1 \;=\; \{\, x \in F : x > 0 \text{ in } \mathbb{R} \,\}, \qquad P_2 \;=\; \{\, x \in F : \sigma(x) \in P_1 \,\},$

and $\sigma$ exchanges them. They differ: $u \in P_1$ and $u \notin P_2$. In the second order $\sqrt 2$ is negative, and indeed lies below every positive rational, while $-\sqrt 2$ is positive; the rationals themselves are ordered the same way in both.

The point of the example is that an order is extra structure on a field, not a property of it: the same field is an ordered field in two inequivalent ways, and no algebraic property of $F$ can distinguish $u$ from $-u$.

• Two orders, one field, and no way to tell them apart algebraically. The automorphism $\sigma$ carries $(F,P_1)$ isomorphically onto $(F,P_2)$ as an ordered field, so the two ordered fields are isomorphic even though the two orders on the underlying $F$ are different subsets. That is the precise sense in which an order is not determined by the field: what is determined here is the order up to isomorphism, not the order itself.

• Contrast with $\mathbb{Q}$ and with $\mathbb{R}$, each of which carries exactly one order. For $\mathbb{Q}$ this is step 1.2: every rational is a quotient of canonical naturals, so its sign is forced. For $\mathbb{R}$ it is Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$: the positives are exactly the nonzero squares, and the squares are fixed by the field structure alone. $\mathbb{Q}(\sqrt 2)$ sits between the two and has room for exactly two, because $2$ acquires a square root while $F$ still has elements that are not squares.

• What decides an order on $F$ is a single bit, the sign of $u$, after which every other comparison reduces to a comparison of rationals. That is also why there are exactly two and not more: the sign of $u$ is the only free choice, and both of its values are realised.

Let $K = \mathbb{R}((t^{-1}))$ be the field of formal Laurent series in $t^{-1}$ over $\mathbb{R}$ (The formal Laurent series $\mathbb{R}((t^{-1}))$: support bounded below, valuation, leading coefficient), ordered by the sign of the leading coefficient ($\mathbb{R}((t^{-1}))$ is an ordered field, ordered by the sign of the leading coefficient). This example assembles, in one place and against the five properties of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, what the field does and does not satisfy:

$K$ is therefore the witness for FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property, and a worked illustration of how far apart the two things called "completeness" can be. A concrete convergent Cauchy sequence is exhibited at the end.

• The one-line reason. Comparison in $K$ looks only at the first coefficient at which two elements differ, so $t$ is bigger than every real constant and $t^{-1}$ is smaller than every positive real constant. The naturals are therefore bounded, which kills (LUB), (BW) and (MCT) at a stroke. Meanwhile a Cauchy sequence in $K$ must have each of its coefficients eventually constant, and reading off those eventual values builds the limit; nothing about the naturals being cofinal is needed for that.

• Why the limit above is not a sum. The notation $\sum_{k \ge 0} t^{-k}$ for $L$ is a name for a function, not an infinite sum (The formal Laurent series $\mathbb{R}((t^{-1}))$: support bounded below, valuation, leading coefficient). What step 2.3 proves is a genuine limit in the order of $K$, and it happens to agree with that notation; no notion of convergence is presupposed by the notation itself.

• What this example does not give. It says nothing about $\mathbb{R}(t)$, the other non-Archimedean field in this library (The rational function field $\mathbb{R}(t)$ ordered by the eventual sign is an ordered field, worked out), which is neither Cauchy complete nor nested-interval complete and cannot replace $K$ in any of these roles.

• Uniqueness of the complete ordered field is untouched. $K$ is not a complete ordered field, so it is no counterexample to that uniqueness; it is a counterexample only to the habit of calling (CC) completeness.

Let $(s_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$, the unique sequence of reals with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$, usually written $s_k = (-1)^k$. Its Cesaro means (The Cesaro means $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$-summability) are

$\sigma_n \;=\; \frac{s_0 + \dots + s_n}{n+1} \;=\; \begin{cases} \dfrac{1}{n+1} & n \text{ even},\\[4pt] 0 & n \text{ odd},\end{cases}$

so the first few values are

$\sigma_0 = 1,\quad \sigma_1 = 0,\quad \sigma_2 = \tfrac13,\quad \sigma_3 = 0,\quad \sigma_4 = \tfrac15,\quad \sigma_5 = 0,\ \dots$

and $\lim_n \sigma_n = 0$, while $(s_k)$ does not converge at all. So $(s_k)$ is $(C,1)$-summable to $0$ and divergent: it is the standard witness that $(C,1)$-summability is strictly weaker than convergence, and the one used in FALSE: if the Cesaro means of a sequence converge then the sequence converges.

The value $0$ is the one an average ought to give, since the sequence spends half its indices at $1$ and half at $-1$; the classical way to say this is that the series $1 - 1 + 1 - 1 + \dots$ has Cesaro sum $\tfrac12$, that being the Cesaro limit of its partial sums rather than of its terms.

• The means converge but are not monotone, and they are not even eventually of one shape: they alternate between $0$ and a positive value shrinking like $1/(n+1)$. Convergence of a Cesaro transform therefore carries no monotonicity information, which is another way of seeing that the transform loses the oscillation rather than damping it.

• Where the $1/2$ comes from. The classical assertion "$1 - 1 + 1 - 1 + \dots = 1/2$" is about the partial sums $S_m$, which are $0, 1, 0, 1, \dots$; their Cesaro means tend to $1/2$. This library has no theory of series yet, so nothing above asserts it; the sequence averaged here is $(s_k)$ itself, whose means tend to $0$.

• This is not a failure of the Cesaro matrix. That matrix is regular (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem): it never changes a limit that exists. What it does here is assign a value where no limit exists, which is exactly what a summability method is for.

Fix a natural number $p \ge 1$ and put

$a_n \;:=\; \sum_{k=0}^{n} k^p \;=\; 1^p + 2^p + \dots + n^p, \qquad b_n \;:=\; n^{p+1},$

the first equality because $0^p = 0$ for $p \ge 1$ (Integer powers $a^m$). Then $(b_n)$ is strictly increasing and unbounded with $b_0 = 0$ and $b_n > 0$ for $n \ge 1$, so Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ applies with $K_0 = 1$, and

$\lim_{n} \frac{1^p + 2^p + \dots + n^p}{n^{p+1}} \;=\; \frac{1}{p+1},$

the limit being taken over the indices $n \ge 1$, where the quotient is defined. For $p = 1$ this is

$\lim_{n} \frac{1 + 2 + \dots + n}{n^2} \;=\; \frac12 .$

No closed form for $a_n$ is used. That is the point of the example: the difference quotient of Stolz-Cesaro replaces a summation formula by a single algebraic identity, the factorisation of $(n+1)^{p+1} - n^{p+1}$.

• Why the limit is taken from $n = 1$. $b_0 = 0$, so $a_0/b_0$ does not denote anything, and Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ is stated for the tail exactly for this reason. Nothing is lost: convergence is a property of a tail (Convergence depends only on the tail).

• The closed form is available and is not needed. For $p = 1$ one has $a_n = n(n+1)/2$, and dividing by $n^2$ gives the limit $1/2$ directly. For general $p$ the closed form is Faulhaber's formula, which this library does not prove; the difference quotient sidesteps it entirely, and that is the practical content of Stolz-Cesaro.

• A sanity check on the answer. The quotient compares a sum of $n$ terms, the largest of which is $n^p$, with $n^{p+1} = n \cdot n^p$, so the limit must lie in $[0,1]$; and the terms $k^p$ grow, so the sum should be a definite fraction of the largest term times $n$. The fraction is $1/(p+1)$, which is what an integral comparison would also predict. No such comparison is used above.