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The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis

Statement

Let $F$ be an ordered field with the monotone convergence property (MCT) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ is Archimedean (Archimedean ordered field).

So (MCT), like (BW) and like (LUB), carries the Archimedean property on its own, and the hypothesis attached to (CC) in Cauchy completeness plus the Archimedean property imply the monotone convergence property need not be attached here. This is what lets Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it sort the five properties into those that do and those that do not.

Facts & Assumptions

Given: An ordered field $F$ with (MCT).

[L1]

The property (MCT): every nondecreasing sequence in $F$ that is bounded above converges in $F$ (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L2]

Sequences in an ordered field: a sequence is a function $N \to F$ ; it is nondecreasing when $x_{j} \leq x_{k}$ for all $j \leq k$ ; convergence and Cauchyness in $F$ are as fixed there (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: $F$ is Archimedean when for every $x \in F$ there is a natural $n$ with $x < n \cdot 1_{F}$ , where $0 \cdot 1_{F} = 0$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ (Archimedean ordered field).

[L4]

Canonical naturals: $n \cdot 1_{F} > 0$ for $n \geq 1$ and $n \mapsto n \cdot 1_{F}$ is strictly increasing on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[L5]

A sequence converging in $F$ is Cauchy in $F$ (clause 3 of Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges).

[L6]

Order arithmetic: $0 < 1_{F}$ (The multiplicative identity is positive); the order is total, so the failure of $x < y$ is $y \leq x$ ; adding a constant preserves the order (Order is preserved by adding a constant and by adding inequalities, Ordered field); and $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value). Here Order is preserved by adding a constant and by adding inequalities states the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

Proof

technique · contradiction

1.1

Suppose $F$ has (MCT) and is not Archimedean; then there is $x \in F$ with $n \cdot 1_{F} \leq x$ for every $n \in N$ .

L3L6assume-contra

1.2

Let $(y_{k})$ be the sequence $y_{k} : = k \cdot 1_{F}$ in $F$ , so $y_{0} = 0$ and $y_{k + 1} = y_{k} + 1_{F}$ ; it is nondecreasing, since $0 = y_{0} < y_{n}$ for $n \geq 1$ and $j \mapsto j \cdot 1_{F}$ is strictly increasing on the positive naturals.

L2L3L4

2.1

$(y_{k})$ is bounded above by $x$ , so (MCT) makes it converge in $F$ to some $L$ .

step 1.1step 1.2L1L2

3.1

Being convergent, $(y_{k})$ is Cauchy in $F$ , so, $1_{F}$ being positive, there is $N \in N$ with $∣ y_{k} - y_{l} ∣ < 1_{F}$ for all $k, l \geq N$ .

step 2.1L2L5L6

4.1

But $y_{N + 1} - y_{N} = 1_{F} > 0$ , so $∣ y_{N + 1} - y_{N} ∣ = 1_{F}$ , which is not $< 1_{F}$ ; this contradicts step 3.1.

step 1.2step 3.1L6

5.1

The assumption of step 1.1 is therefore untenable, and an ordered field with (MCT) is Archimedean.

step 4.1discharge-contradiction∎

Remarks

Why this item exists. Without it, the natural reading of the equivalence theorem would attach an Archimedean hypothesis to (MCT) as well, and Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it would answer its own question wrongly. With it the answer is clean: (LUB), (BW) and (MCT) each imply the Archimedean property, while (NIP) and (CC) do not (FALSE: the nested interval property alone implies the least-upper-bound property, FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property).
The witness is the same sequence as in Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis, the canonical naturals, but the two arguments use different failures of it. There the sequence is bounded and has no convergent subsequence; here it is nondecreasing and bounded above and has no limit. Neither argument implies the other, because neither (BW) nor (MCT) is assumed in the other's proof.
The gap of exactly $1_{F}$ between consecutive terms is what does the work, and it is available in every ordered field: $1_{F} > 0$ by The multiplicative identity is positive, and no smallness of $1_{F}$ relative to $x$ is possible, since the Cauchy condition is tested at the threshold $1_{F}$ itself.

Depends on

Used by

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29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. F. Hall, Completeness of Ordered Fields (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
Monotone convergence theorem (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §1.2 and §2.4 (standard reference, not scraped)