LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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An ordered field with the least-upper-bound property has the nested interval property and is Archimedean

Statement

Let $F$ be an ordered field with the least-upper-bound property (LUB) 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);
$F$ has the nested interval property (NIP).

The intersection point produced in claim 2 is the supremum of the left endpoints, and the proof does not use the hypothesis that the lengths tend to $0$ : an ordered field with (LUB) satisfies the unrestricted nested interval property, of which (NIP) as defined is a special case.

Facts & Assumptions

Given: An ordered field $F$ with the least-upper-bound property, and a nested sequence $(I_{k})_{k \in N}$ of closed intervals $I_{k} = [a_{k}, b_{k}]_{F}$ of $F$ , so that $a_{k} \leq b_{k}$ for every $k$ and $I_{k + 1} \subseteq I_{k}$ for every $k$ .

[L1]

Least upper bounds: every nonempty $S \subseteq F$ bounded above has a least upper bound $\sup S \in F$ ; a least upper bound is an upper bound and is $\leq$ every upper bound (Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound).

[L2]

Every complete ordered field is Archimedean (Every complete ordered field is Archimedean).

[L3]

The properties (LUB), (NIP) and the Archimedean property, as fixed in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and Archimedean ordered field; (LUB) for $F$ is by definition the statement that $F$ is a complete ordered field (Complete ordered field (least-upper-bound property)).

[L4]

Closed intervals and nesting in $F$ : $[a, b]_{F} = {x \in F : a \leq x \leq b}$ for $a \leq b$ , and $(I_{k})$ is nested when $I_{k + 1} \subseteq I_{k}$ for every $k$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L5]

The order of $F$ is total and transitive (Ordered field).

[L6]

Induction principle on $N$ (The principle of mathematical induction), and the order on $N$ is total, so of any two indices one is the larger ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

Having (LUB) is by definition being a complete ordered field, so $F$ is a complete ordered field.

L1L3

1.2

For every $k$ : $a_{k + 1}$ and $b_{k + 1}$ lie in $I_{k + 1}$ , hence in $I_{k}$ , so $a_{k} \leq a_{k + 1} \leq b_{k + 1} \leq b_{k}$ .

L4L5

2.1

$F$ is Archimedean, which is claim 1.

step 1.1L2

2.2

By induction on the difference of the indices, $a_{j} \leq a_{m}$ and $b_{m} \leq b_{j}$ whenever $j \leq m$ .

step 1.2L5L6

3.1

For all $j, l \in N$ one has $a_{j} \leq b_{l}$ : letting $m$ be the larger of $j$ and $l$ , $a_{j} \leq a_{m} \leq b_{m} \leq b_{l}$ .

step 1.2step 2.2L5L6

4.1

The set $A : = {a_{k} : k \in N}$ is nonempty and is bounded above by $b_{0}$ , so $c : = \sup A$ exists in $F$ .

step 3.1L1

5.1

For every $k$ : $a_{k} \leq c$ because $c$ is an upper bound of $A$ ; and $c \leq b_{k}$ because $b_{k}$ is an upper bound of $A$ by step 3.1 while $c$ is the least such.

step 3.1step 4.1L1

6.1

So $c \in [a_{k}, b_{k}]_{F}$ for every $k$ , the intersection of the $I_{k}$ is nonempty, and $F$ has (NIP), which with step 2.1 gives both claims.

step 2.1step 5.1L3L4∎

Remarks

Where the lengths would be used. They are not used at all above. Their role is to force the intersection to be a single point: if the lengths tend to $0$ and $c, c^{'}$ both lie in every $I_{k}$ then $∣ c - c^{'} ∣ \leq b_{k} - a_{k}$ for every $k$ , so $∣ c - c^{'} ∣$ is below every positive element of $F$ and therefore $0$ . Uniqueness is not part of (NIP) as defined in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and is not needed anywhere on this page.
The converse of claim 1 fails, and that is the point of two items later on this page. FALSE: the nested interval property alone implies the least-upper-bound property shows that (NIP) does not imply (LUB), and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property shows the same for (CC); in both the witness is a non-Archimedean field, so neither carries the Archimedean property that (LUB) carries here.

Depends on

Used by

Dependency tree · two levels

33 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)
Nested intervals (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)