Every complete ordered field is Archimedean

Depends on

• Complete ordered field (least-upper-bound property)
• Canonical naturals are positive and strictly increasing
• The multiplicative identity is positive
• Ordered field

Used by

• A Lebesgue measurable subgroup of (ℝⁿ,+) of positive measure is all of ℝⁿ Corollary
• Brownian paths are locally Holder below one half Corollary
• For every ε > 0 in a complete ordered field there is a natural n ≥ 1 with 1/n < ε Corollary
• Integral test as an equivalence with an improper integral Corollary
• ℝ((t⁻¹)) does not have the least-upper-bound property; its canonical naturals have no supremum Corollary
• ℝⁿ is locally compact and σ-compact Corollary
• ⋂ₖ (-1/k, 1/k) = {0} is not open Counterexample
• A complete domain is necessary for sequential uniform boundedness Counterexample
• A function differentiable on [0,1] whose derivative is unbounded, hence not Riemann integrable Counterexample
• A nonnegative function can have both iterated integrals zero and no double Riemann integral Counterexample
• A sequence with limsup = +∞: the greatest subsequential limit exists only in ℝ̄ Counterexample
• A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one Counterexample
• aₖ = (-1)ᵏ, bₖ = k have aₖ/bₖ → 0 while the difference quotient oscillates, so Stolz-Cesaro has no converse Counterexample
• An unbounded set has no supremum: the naturals inside ℝ Counterexample
• Collapsing the set of naturals inside ℝ to a point gives a quotient of ℝ that is not locally compact at the collapsed point Counterexample
• Continuous fₙ → 0 pointwise on [0,1] with ∫₀¹ fₙ = 1 for every n Counterexample
• Finite target bounds do not supply an infinite target bound Counterexample
• For the Dirichlet function every uniform partition with rational tags gives Riemann sum 1, so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence Counterexample
• ℕ with the discrete metric is bounded and is not totally bounded Counterexample
• Not every ordered field is Archimedean Counterexample
• Null times divergent has no rule: xₖ = 1/k with yₖ = ck gives product limit c, and with yₖ = k² gives divergence Counterexample
• On (0,∞) the metrics |x-y| and |1/x - 1/y| have the same topology and are not uniformly equivalent Counterexample
• On (0,∞) the metrics |x-y| and |1/x - 1/y| share their topology and not their Cauchy sequences Counterexample
• On (0,1) the identity is bounded with no greatest value and x ↦ 1/x is continuous and unbounded, so the extreme value theorem needs compactness and not merely boundedness of the domain Counterexample
• On ℕ with d(m,n) = 1 + 1/(m+n) for m ≠ n the sets {n, n+1, …} are nested, closed, bounded and complete with empty intersection Counterexample
• On ℝ the metrics |x-y| and min(|x-y|,1) are uniformly but not Lipschitz equivalent Counterexample
• On the positive integers the metrics |m-n| and |1/m - 1/n| both induce the discrete topology, and only the first is complete Counterexample
• ℝⁿ is closed and unbounded and is not compact for n≥1 Counterexample
• The cover {(1/k, 1)} of (0,1) has no finite subcover, so (0,1) is not compact Counterexample
• The cover of (0,1) by the intervals (1/(k+2), 1) has no Lebesgue number, so the Lebesgue number lemma needs compactness Counterexample
• The identity on (0,1) is bounded with no greatest value, and on [0,∞) it is continuous and unbounded Counterexample
• The nested closed unbounded sets [k, ∞) have empty intersection, so boundedness cannot be dropped Counterexample
• The nested open intervals (0, 1/k) have empty intersection Counterexample
• The open interval (0,1) is totally bounded and not compact, the cover by the intervals (1/(k+2), 1) having no finite subcover Counterexample
• The sequence 1, 1, 2, 1, 3, 1, 4, … is unbounded and has a convergent subsequence Counterexample
• The unrestricted nested interval property fails in ℝ((t⁻¹)) Counterexample
• x ↦ √x is a uniformly continuous bijection of [0,∞) onto itself whose inverse x ↦ x² is not uniformly continuous Counterexample
• x ↦ √x on (0,1] is differentiable with unbounded derivative and is not Lipschitz there, so the boundedness hypothesis in the Lipschitz corollary cannot be dropped Counterexample
• x ↦ 1/x is continuous on (0,1) and not uniformly continuous there, the pairs 1/(k+2) and 1/(k+3) defeating every δ Counterexample
• x ↦ 1/x is continuous on (0,1) and not uniformly continuous, so Heine-Cantor needs compactness of the domain Counterexample

…and 152 more results.

Dependency tree · two levels

11 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

• W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
• T. Tao, Analysis I, 3rd ed. (standard reference, not scraped)
• Neil Donaldson, Math 140A notes: Completeness and the Archimedean property (standard reference, not scraped)