Complete ordered field (least-upper-bound property)

Depends on

• Ordered field

Used by

• Abel's test for improper integrals Corollary
• Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫ₐᵇ f = G(b)-G(a) for any primitive G Corollary
• Every nondegenerate interval of ℝ is uncountable Corollary
• For every ε > 0 in a complete ordered field there is a natural n ≥ 1 with 1/n < ε Corollary
• If f,g are integrable on [a,b] then so are | f|, f², fg, max(f,g) and min(f,g), and |∫ₐᵇ f| ≤ ∫ₐᵇ| f| Corollary
• ℝ((t⁻¹)) does not have the least-upper-bound property; its canonical naturals have no supremum Corollary
• Stolz-Cesaro, 0/0 form: if bₖ is strictly decreasing to 0, aₖ → 0, and the difference quotient converges, then aₖ/bₖ converges to the same value Corollary
• The Cauchy-sequence reals have the least-upper-bound property Corollary
• The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem Corollary
• The irrationals are uncountable Corollary
• [0,1) is neither open nor closed in ℝ Counterexample
• {0} ∪ [1,2] is closed, has an isolated point, and is not perfect Counterexample
• ⋂ₖ (-1/k, 1/k) = {0} is not open Counterexample
• 1/4 lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it Counterexample
• A function differentiable on [0,1] whose derivative is unbounded, hence not Riemann integrable Counterexample
• A function that is not Riemann integrable although | f| is 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
• A supremum need not belong to its set: sup(0,1) = 1 ∉ (0,1) 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 f and integrable sign-changing g with ∫ₐᵇ fg ≠ f(ξ)∫ₐᵇ g for every ξ Counterexample
• Continuous fₙ → 0 pointwise on [0,1] with ∫₀¹ fₙ = 1 for every n 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
• In {0} ∪ [1,2] with the metric of ℝ, the closure of B(0,1) = {0} is {0} while the closed ball is {0,1} Counterexample
• Integrable φ and integrable f with φ∘ f not integrable: the order of the hypotheses in the composition theorem cannot be reversed 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 (0,1) 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 a closed interval of ℚ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property 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
• Over ℚ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail Counterexample
• ℚ ∩ [0,2] is bounded and disconnected, so being an interval of ℚ is not enough Counterexample
• ℝ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions Counterexample
• Refuted: convergence uniformly on every compact subset of ℝ implies uniform convergence. The maps x ↦ x/(n+1) separate the two Counterexample
• The cover {(1/k, 1)} of (0,1) has no finite subcover, so (0,1) is not compact Counterexample
• The Dirichlet function on [0,1] has lower Darboux integral 0 and upper Darboux integral 1, so it is bounded and not Riemann integrable Counterexample
• The empty set is bounded and has no supremum Counterexample

…and 291 more results.

Dependency tree · two levels

2 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)
• M. Spivak, Calculus, 4th ed., Ch. 8 (standard reference, not scraped)
• University of Wisconsin Math 521 notes: Real analysis (standard reference, not scraped)