How statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Both and are dense in , and every nonempty open subset of is uncountable
Statement
Write for the image of in under the canonical embedding (The rationals embed densely in the reals), the set usually written once the identification is made, and put for the irrationals. Then:
- is dense in , that is, (Limit point, isolated point, adherent point, derived set, and dense subset of );
- is dense in ;
- every nonempty open subset of is uncountable (Finite, countably infinite, countable, uncountable).
Claim 2 is not a symmetry of claim 1: the rationals are dense because they are constructed to approximate, whereas the irrationals are dense because there are too many points in any interval for a countable set to exhaust it, which is why claim 3 is proved alongside and used for it.
Facts & Assumptions
Given: The canonical embedding of into , its image , and the complement .
is the set of points every neighbourhood of which meets ; is dense in when (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Limit point, isolated point, adherent point, derived set, and dense subset of ).
is open when every admits with (Open subset of (every point has a neighbourhood inside it), closed subset (complement open), and clopen).
Strictly between any two reals lies an element of , and is injective (The rationals embed densely in the reals).
( is countably infinite); an injection is a bijection onto its image, and is symmetric and transitive (Injection, surjection, bijection, Equinumerous sets, and ).
Every subset of an at most countable set is at most countable, and uncountable means not at most countable (Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).
For the interval is uncountable (Every nondegenerate interval of is uncountable).
Proof
is dense: let and let be real; by [L2] one has , so [L4] supplies with , that is . Every real is therefore an adherent point of and claim 1 follows from [L1].
is at most countable: the embedding is an injection of with image , hence a bijection onto it, so .
For all reals the interval is uncountable.
For all reals the interval contains an irrational: if it did not, then , so would be a subset of an at most countable set by step 1.2 and hence at most countable by [L6], contradicting step 1.3. So some lies in .
Every nonempty open is uncountable: fix and, by [L3], a real with ; by [L2] the set is the interval with , hence uncountable by step 1.3. Were at most countable, its subset would be at most countable by [L6], which it is not; so is uncountable, which is claim 3.
is dense: let and let be real; applying step 2.1 with and gives , which is by [L2]. Every real is therefore an adherent point of , so by [L1], which is claim 2.
Claims 1, 2 and 3 are steps 1.1, 3.1 and 2.2, so both and its complement are dense in and every nonempty open subset of is uncountable.
Remarks
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Two dense sets can be disjoint. and partition and both are dense, so density says nothing about size: one of them is countable and the other is not (The irrationals are uncountable). What density does say is that neither has interior: a set whose complement is dense has empty interior, which is the computation carried out for in has closure , empty interior, and boundary ↗.
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Claim 3 is a statement about open sets, not about intervals. It follows from the uncountability of intervals (Every nondegenerate interval of is uncountable) only because openness supplies an interval inside the set at each of its points. A nonempty set with empty interior can perfectly well be countable, as shows.
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An explicit irrational is not produced here. Step 2.1 is a counting argument and exhibits nothing. The library does exhibit one separately, (Square roots exist: a unique with ; the positives are , FALSE: some rational number squares to 2), and an explicit irrational in a given interval can be built from it as for suitable rationals in the interval; that route is longer and is not the one taken above.
Depends on
- The rationals embed densely in the reals
- Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$
- Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen
- Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length
- The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$
- The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points
- Every nondegenerate interval of $\mathbb{R}$ is uncountable
- Every subset of an at most countable set is at most countable
- Finite, countably infinite, countable, uncountable
- $\mathbb{Q}$ is countably infinite
- Equinumerous sets, $A \approx B$ and $A \preceq B$
- Injection, surjection, bijection
Used by
- A finite-dimensional normed subspace is closed Corollary
- c₀ is not isomorphic to a dual space Corollary
- ℚ is F_σ, meager and not G_δ, while the irrationals are G_δ, residual and not F_σ Corollary
- A closed subspace of ell-infinity that is not complemented Counterexample
- A function that is not Riemann integrable although | f| is Counterexample
- A nonnegative function can have both iterated integrals zero and no double Riemann integral 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
- ℚ ∩ [0,1] has measure zero and not content zero, although it is bounded Counterexample
- ℚ is dense in ℝ and has measure zero Counterexample
- ℚ∩[0,1] is Lebesgue null and has Jordan outer content one 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 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 Dirichlet right-hand side gives a first-order equation with no solution Counterexample
- The indicator of ℚ has a limit at no point of ℝ Counterexample
- The indicator of ℚ is continuous at no point of ℝ Counterexample
- The rational points of [0,1]² form a bounded null set that is not Jordan measurable 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
- Whitehead theorem fails without CW type Counterexample
- Complex Lp classes and Euclidean test-function conventions Definition
- The circle, rotations and the doubling map Definition
- The Dirichlet function 1_ℚ, and Thomae's function t with t(x) = 1/q at a rational x = p/q in lowest terms with q ≥ 1 and t(x) = 0 at every irrational x Definition
- A bounded function on ℝ with no local maximum and no local minimum at any point, upper semicontinuous at no point and lower semicontinuous at no point: compose the Hamel coefficient with a strictly increasing injection of ℝ into (0,1) Example
- A bounded increasing integrand discontinuous at every rational has an integral function nondifferentiable at every rational in (0,1) Example
- An additive f : ℝ → ℝ that is not x ↦ cx: the coefficient of one fixed Hamel basis vector. It is unbounded above and below on every nondegenerate interval, its graph is dense in ℝ², and every nonempty level set is dense in ℝ Example
- An open dense set of measure less than 1 is the monotone L¹-limit of Riemann integrable indicators, but its indicator is not Riemann integrable Example
- Multiplication operators: domain, spectral measure and spectrum Example
- ℚ as a subspace of ℝ: every component is a single point, no point is isolated, and the space is not locally connected anywhere Example
- ℚ has closure ℝ, empty interior, and boundary ℝ Example
- ℝ and ℚ are σ-compact, and Lindel"of assuming countable choice; ℝ is locally compact and ℚ is nowhere locally compact Example
- The Dirichlet function is Borel measurable and nowhere continuous Example
- The Dirichlet function is the pointwise limit of a sequence of Baire class one functions and is itself not Baire class one, so the Baire hierarchy on [0,1] is already strict at the first level Example
- The Dirichlet function satisfies Lusin's conclusion without being continuous anywhere Example
- The function equal to q at a rational p/q in lowest terms and to 0 at every irrational is finite at every point and unbounded on every nondegenerate interval Example
- The Gauss map preserves Gauss measure Example
- The indicator of the irrationals is Henstock–Kurzweil integrable with integral 1 Example
- The rationals are Borel and F-sigma but neither open nor closed nor G-delta Example
- The rationals are dense but have dimension zero Example
- Thomae's function computed: t(1/2) = 1/2, t(2/3) = 1/3, t(m) = 1 at every integer m, t(x) = 0 at every irrational, and ωₜ(c) = t(c) at every real c Example
- Thomae's function is Riemann integrable on [0,1] with integral 0: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is 0 Example
- Thomae's integrand is discontinuous at every rational, yet its integral function is identically zero and differentiable everywhere Example
…and 36 more results.
Dependency tree · two levels
63 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
- Dense set (Wikipedia) (standard reference, not scraped)
- Countable set (Wikipedia) (standard reference, not scraped)
- W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 and Ch. 2 (standard reference, not scraped)
- J. Lebl, Basic Analysis I, §1.2 (standard reference, not scraped)
- J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)