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A nonempty set is at most countable iff it is a surjective image of
Statement
Let be a nonempty set. Then is at most countable (Finite, countably infinite, countable, uncountable) if and only if there is a surjection (Injection, surjection, bijection).
Moreover, from any such surjection an injection is obtained explicitly, without any choice, by
This is the working form of countability used everywhere below: to prove a nonempty set countable it suffices to list its elements, repetitions and all.
No choice principle is used. The backward direction is where an appeal to choice would be natural ("for each pick some with ") and it is avoided outright, because is canonical: every nonempty set of naturals has a least element (The well-ordering principle), so is determined by and alone.
Facts & Assumptions
Given: A nonempty set . For and a function write .
is at most countable when for some or ; holds only for (Finite, countably infinite, countable, uncountable, The natural numbers (von Neumann)).
Bijections, injections, surjections, images and the symmetry and transitivity of ; an injection is a bijection onto its image (Injection, surjection, bijection, Equinumerous sets, and ).
Well-ordering: every nonempty subset of has a least element (The well-ordering principle).
Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).
For naturals, , so a natural number is the set of naturals below it; in particular whenever (On the order is membership: , proved earlier on this page from the additive order of Order on the natural numbers on the von Neumann naturals of The natural numbers (von Neumann)).
Proof
For the forward implication assume is at most countable; since we have , or for some with , and in either case fix a bijection from , respectively from , onto .
For the converse implication assume a surjection is given.
If is defined on it is itself a surjection ; if is defined on , then by [L5] and the function with for and for is a surjection, since every element of is for some . In both cases a surjection exists.
For each the set is a nonempty subset of , because is surjective, so [L3] provides its least element and defines a function ; no selection is made, since the least element is uniquely determined.
is injective: if then and , because and , so .
Hence is a bijection of onto , so ; the subset of the at most countable set is at most countable by [L4], and transitivity of transfers this to .
The forward implication is step 2.1 and the converse is step 4.1, so for nonempty countability and the existence of a surjection are equivalent, with of step 2.2 the promised injection .
Remarks
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The hypothesis cannot be dropped in the forward direction: is finite, hence at most countable, but no function exists at all. The converse direction needs no such hypothesis, since a surjection onto already forces .
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Combining the two directions: a nonempty is at most countable if and only if (Equinumerous sets, and ). The forward direction of that reformulation is immediate, and the backward direction is step 4.1.
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The lemma is what licenses the informal phrase "enumerate as , possibly with repetitions". Repetitions are exactly what distinguishes a surjection from a bijection, and allowing them is what makes the criterion easy to apply: the enumerations built in A product of two at most countable sets is at most countable and Countable unions of at most countable sets, assuming repeat.
Depends on
- Finite, countably infinite, countable, uncountable
- The well-ordering principle
- Injection, surjection, bijection
- Every subset of an at most countable set is at most countable
- Equinumerous sets, $A \approx B$ and $A \preceq B$
- The natural numbers $\mathbb{N}$ (von Neumann)
- Order on the natural numbers
- On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$
Used by
- Every nondegenerate interval of ℝ is uncountable Corollary
- Every subspace of a metrizable space is metrizable and every subspace of a first countable space is first countable, the metric case being the subspace metric already identified with the subspace topology Corollary
- The irrationals are uncountable Corollary
- 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
- Refuted, assuming countable choice: every Hausdorff space built from ordinal spaces is normal. The deleted Tychonoff plank ((ω₁ + 1) × (ω + 1)) ∖ {(ω₁, ω)} is Hausdorff and not normal Counterexample
- The antidiagonal {(x,-x)} is an uncountable discrete subspace of the Sorgenfrey plane, so having a countable dense subset is not a hereditary property Counterexample
- Countably compact, Lindel"of, sequentially compact, limit point compact and σ-compact spaces, and relatively compact subsets Definition
- Countably compact, sequentially compact and limit point compact metric spaces Definition
- The Axiom of Countable Choice (AC_ω) Definition
- A bounded nondecreasing f : ℝ → ℝ whose set of discontinuities is exactly ℚ, obtained from the prescribed-jump construction applied to one fixed enumeration of the rationals Example
- Assuming countable choice, a strictly increasing ω-sequence of countable ordinals has a countable supremum, which is a countable limit ordinal below ω₁; the instance supₙ ω·(n+1) = ω² needs no choice Example
- Assuming countable choice, cf(ℵ_ω₁) = ℵ₁, so singular does not mean of countable cofinality Example
- Baire category gives a third proof that ℝ is uncountable Example
- Froda's countable bound is attained: a bounded nondecreasing function on ℝ discontinuous exactly at the points 1 - 1/(k+1) for k ∈ ℕ, an infinite discontinuity set inside a bounded interval Example
- In the cocountable topology on ℝ the closed sets are the countable sets and ℝ, and a sequence converges iff it is eventually constant Example
- ℚ is covered by open intervals of total length ε, for every ε > 0 Example
- ℝ and ℚ are σ-compact, and Lindel"of assuming countable choice; ℝ is locally compact and ℚ is nowhere locally compact Example
- ℝ with the half-open intervals [a,b) as a basis is not compact and, assuming the Axiom of Countable Choice, is Lindel"of, while its square is not Lindel"of, the antidiagonal being an uncountable closed discrete subspace Example
- The cocountable topology on ℝ is T₁, has unique sequential limits, and is neither Hausdorff nor regular nor normal 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 discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies placed in the connectedness hierarchy Example
- The Sorgenfrey line: ℝ with the half-open intervals [a,b) as a basis is strictly finer than the usual topology, is first countable, has a countable dense subset, and its sequences converge only from the right Example
- ω + 1 as a convergent sequence together with its limit, and, assuming countable choice, [0, ω₁), in which every sequence lies inside an at most countable initial segment Example
- Assuming countable choice, refuted: Lindelöfness is productive False statement
- FALSE: a pointwise limit of a sequence of Riemann integrable functions on [a,b] is Riemann integrable False statement
- FALSE: a sequentially continuous map between topological spaces is continuous False statement
- FALSE: a space in which every sequence has at most one limit is Hausdorff False statement
- FALSE: the Cantor set is countable because only countably many intervals were removed False statement
- Refuted: every separable space is second countable False statement
- A compact metric space has a countable dense subset, by countable choice Lemma
- A countable local base can be chosen open and decreasing Lemma
- Assuming the Axiom of Choice, ℝ has a Hamel basis over ℚ: there is B ⊆ ℝ such that every real is a finite ℚ-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined ℚ-linear coefficient map Lemma
- Every at most countable subset of ℝ has measure zero Lemma
- The balls B(x, 1/n), n ≥ 1, form a countable neighbourhood base at x, so every metric space is first countable Lemma
- Under choice, if |I|>2^ℵ₀, then the Cantor cube 2^I is not separable Lemma
- Every separable space satisfies the countable chain condition Proposition
- The continuum hypothesis, and what this page does not prove Remark
- A product of two at most countable sets is at most countable Theorem
- Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there Theorem
- Assuming countable choice: every at most countable subset of ω₁ is bounded below ω₁, so no at most countable subset of ω₁ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable Theorem
…and 11 more results.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 44 results over 22 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
- J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
- J. Lebl, Basic Analysis: Introduction to Real Analysis, basic set theory (standard reference, not scraped)
- Countable set (Wikipedia) (standard reference, not scraped)
- T. Tao, Analysis I, 3rd ed., §8.1 (standard reference, not scraped)