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Order is preserved by adding a constant and by adding inequalities
Statement
Let be an ordered field (Ordered field) with positive cone , and let .
- Translation invariance. If then .
- Adding inequalities. If and then .
Facts & Assumptions
Given: An ordered field with positive cone , and elements .
For , the relation means (Ordered field).
is closed under addition: if then (axiom O2 of Ordered field).
Proof
Assume ; by the definition of the order this means .
For every the field identities give .
Assume moreover ; by the definition of the order this means .
The field identities give .
Hence , which is exactly , proving claim 1.
Since and , closure under addition gives .
Therefore , which is exactly , proving claim 2.
Depends on
Used by
- Every nondegenerate interval of ℝ is uncountable Corollary
- ℝ((t⁻¹)) has the nested interval property for lengths tending to 0 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 limit inferior is the least subsequential limit in overlineℝ 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 sequence with limsup = +∞: the greatest subsequential limit exists only in overlineℝ 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
- 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 the domain {0} ∪ [1,2] every real is vacuously a limit at 0 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
- 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
- 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
- The identity from the cocountable topology on ℝ to the usual topology is sequentially continuous and not continuous Counterexample
- The indicator of ℚ has a limit at no point of ℝ Counterexample
- The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero Counterexample
- Thomae's function is nonnegative, Riemann integrable on [0,1] with integral 0, and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish Counterexample
- xₖ = (-1)ᵏ, yₖ = (-1)ᵏ⁺¹ give limsup(xₖ + yₖ) = 0 < 2 = limsup xₖ + limsup yₖ Counterexample
- xₖ = 1 + (-1)ᵏ, yₖ = 1 + (-1)ᵏ⁺¹ give limsup(xₖ yₖ) = 0 < 4 Counterexample
- xₖ₊₁ = xₖ + 1/xₖ from x₁ = 1 has strictly decreasing consecutive gaps and diverges, so no uniform c < 1 exists Counterexample
- ℤ and {n + 1/n : n ≥ 2} are disjoint closed subsets of ℝ at distance 0, so the set-to-set distance is not a metric Counterexample
- ℤ is closed and not compact, and (0,1) is bounded and not compact: neither hypothesis of Heine-Borel can be dropped Counterexample
- ψ(1/x) has no limit at 0: two sequences tending to 0 give values constantly 0 and constantly 1/2 Counterexample
- Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space Definition
- Limits at +∞ and -∞, and infinite limits at a point Definition
- The Cantor function on [0,1], defined on the Cantor set through ternary digits and extended constantly across each removed interval Definition
- The Cantor middle-thirds set as the intersection of the sets Cₙ obtained by removing open middle thirds Definition
- The extended real line overlineℝ = ℝ ∪ {-∞, +∞}, its order, and the arithmetic that is left undefined Definition
- The Smith-Volterra-Cantor set: the same construction removing, at stage n ≥ 1, an open middle interval of length 4⁻ⁿ from each of the 2ⁿ⁻¹ remaining intervals Definition
- (-1)ᵏ has liminf = -1 and limsup = 1, so it does not converge Example
- (0,1) + (2,3) = (2,4), with supremum 4 = sup(0,1) + sup(2,3) Example
- (3x² - 1)/(x² + x) → 3 as x → +∞ Example
- ∫₀¹ x² = 1/3, computed from the Darboux definition with uniform partitions and the closed form ∑_k<n k² = n(n-1)(2n-1)/6 Example
…and 130 more results.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 2 results over 2 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 Illinois Chicago notes: Ordered field axioms (standard reference, not scraped)