Inverses of positives are positive, and reciprocation reverses order

Depends on

• Ordered field
• Sign rules for products and monotonicity of multiplication
• The multiplicative identity is positive

Used by

• 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
• Kummer with ζₖ = 1 recovers the ratio test Corollary
• Limit comparison for positive improper integrals 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 a priori bound d(x^*, xₙ) ≤ qⁿ d(x₁,x₀)/(1-q) and the a posteriori bound d(x^*, xₙ₊₁) ≤ q d(xₙ₊₁,xₙ)/(1-q) Corollary
• The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem Corollary
• ∏_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
• 1/x² has no finite Cauchy principal value at zero 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
• 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
• fₖ(x)=xᵏ⁺¹ converges pointwise but not uniformly on [0,1] 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
• In ℝ(t) the rationals are not dense: no rational lies strictly between 0 and 1/t 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 A = ([0,∞) × ℝ) ∪ (ℝ × {0}) the first projection is a quotient map, by the section x ↦ (x,0), and is neither open nor closed 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
• 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 diagonal x ↦ (x,x,…) from ℝ into ℝ^ℕ is continuous for the product topology and not for the box topology Counterexample
• The double sequence (m+1)/(m+n+2) has unequal iterated limits Counterexample
• The hyperbola {(x,y) : xy = 1} is closed in ℝ² and its image under the first projection is ℝ ∖ {0}, which is not closed Counterexample

…and 172 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 7 results over 5 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)
• University of Innsbruck notes: Ordered fields (standard reference, not scraped)