The multiplicative identity is positive

Depends on

• Squares of nonzero elements are positive
• Field

Used by

• Every nondegenerate interval of ℝ is 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 sequence with limsup = +∞: the greatest subsequential limit exists only in ℝ̄ Counterexample
• f(x) = x on [0,1) with f(1) = 0 is differentiable at every point of (0,1) with f' ≡ 1, yet no c satisfies f(1) - f(0) = f'(c), so continuity on the closed interval cannot be dropped from the mean value theorem 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
• In the cocountable topology on ℝ the sequential closure of [0,1] is [0,1] while its closure is all of ℝ 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 ℝ the metrics |x-y| and min(|x-y|,1) are uniformly but not Lipschitz equivalent Counterexample
• On the domain {0} ∪ [1,2] every real is vacuously a limit at 0 Counterexample
• ℚ ∩ [0,2] is bounded and disconnected, so being an interval of ℚ is not enough 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 empty set is bounded and has no supremum Counterexample
• The function equal to 0 off the origin and to 1 at the origin has limit 0 ≠ 1 there Counterexample
• The identity from the cocountable topology on ℝ to the usual topology is sequentially continuous and not continuous Counterexample
• The identity on [0,1] attains its maximum at 1 and its minimum at 0 with derivative 1 at both, so Fermat's theorem genuinely needs the extremum to be at an interior point Counterexample
• The indicator of ℚ has a limit at no point of ℝ Counterexample
• With g ≡ 0 and f equal to 0 off the origin and 1 at it, lim g = 0 and lim_y → 0 f = 0 while f ∘ g ≡ 1 Counterexample
• x ↦ |x| is continuous everywhere and not differentiable at 0: the difference quotient equals 1 on the right and -1 on the left, so the two one-sided limits differ 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
• 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 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
• A positive sequence making all three inequalities of the ratio-to-root chain strict Example
• aₖ = 2^-k + (-1)ᵏ has liminf aₖ₊₁/aₖ = 1/8, limsup aₖ₊₁/aₖ = 2 and lim aₖ^1/k = 1/2 Example
• Baire category gives a third proof that ℝ is uncountable Example

…and 88 more results.

Dependency tree · two levels

4 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: Axioms and Basic Definitions (standard reference, not scraped)