Canonical naturals are positive and strictly increasing

Depends on

• The multiplicative identity is positive
• Order is preserved by adding a constant and by adding inequalities
• Field

Used by

• A uniform derivative bound gives a uniform Taylor remainder bound Corollary
• Raabe is Kummer with ζₖ = k+1: for positive terms, liminf (k+1)(aₖ/aₖ₊₁ - 1) > 1 gives convergence and limsup < 1 gives divergence Corollary
• The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem Corollary
• The Lagrange and Cauchy forms of Taylor's remainder Corollary
• ‖·‖₁ on ℝ² violates the parallelogram law, so no symmetric bilinear form induces it Counterexample
• ∏_j ≥ 0 (1 + (-1)ʲ/√j+2) has partial products tending to 0 although ∑_j ≥ 0 (-1)ʲ/√j+2 converges Counterexample
• ∑ k^-1/2 diverges and ∑ k⁻² converges, and both have root limit exactly 1 Counterexample
• ⋂ₖ (-1/k, 1/k) = {0} is not open Counterexample
• A curve for which the mean value inequality is an equality, showing the constant cannot be improved Counterexample
• A function differentiable on [0,1] whose derivative is unbounded, hence not Riemann integrable Counterexample
• A nonnegative non-monotone sequence for which ∑ aₖ and ∑ 2ᵏ a_2ᵏ behave differently 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 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
• Continuous triangular spikes on [0,1] converge pointwise to zero but not uniformly when monotonicity is absent Counterexample
• Dini's theorem fails for discontinuous approximants: shrinking interval indicators decrease pointwise to zero but not uniformly Counterexample
• Dini's theorem fails on [0,∞): x/(ι(k+1)+x) decreases pointwise to zero but not uniformly Counterexample
• f(t) = (t², t³) on [0,1]: no ξ satisfies f(1)-f(0) = f'(ξ) Counterexample
• fₖ(x)=xᵏ⁺¹ converges pointwise but not uniformly on [0,1] 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
• g(x,y) = xy/(x²+y²), extended by g(0,0)=0, is continuous in each variable separately and not continuous at the origin 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 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
• 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
• Over ℚ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail Counterexample
• ℝ covered by its closed singletons: every restriction of the indicator of {0} is continuous and the map is not, so the closed pasting lemma needs finiteness Counterexample
• Refuted: a pointwise bounded family of continuous functions is equicontinuous. The spikes are bounded by 1 everywhere and are not equicontinuous at 0 Counterexample
• Refuted: C(X,Y) is closed in the topology of pointwise convergence. The ramps on [0,1] converge pointwise to a discontinuous limit Counterexample
• Refuted: convergence uniformly on every compact subset of ℝ implies uniform convergence. The maps x ↦ x/(n+1) separate the two Counterexample
• Shrinking rectangles converge pointwise to zero while every integral equals one Counterexample
• The Cauchy product of ∑_k ≥ 0 (-1)ᵏ/√k+1 with itself has |cₙ| ≥ 1 for every n, so it diverges Counterexample
• The cover {(1/k, 1)} of (0,1) has no finite subcover, so (0,1) is not compact Counterexample

…and 191 more results.

Dependency tree · two levels

5 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. 1 (standard reference, not scraped)
• Elias Zakon, Mathematical Analysis: Natural Numbers and Induction (standard reference, not scraped)