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A nonempty set is at most countable iff it is a surjective image of $N$

Statement

Let $A$ be a nonempty set. Then $A$ is at most countable (Finite, countably infinite, countable, uncountable) if and only if there is a surjection $s : N \to A$ (Injection, surjection, bijection).

Moreover, from any such surjection an injection $A \to N$ is obtained explicitly, without any choice, by

$j (a) : = \min {k \in N : s (k) = a} .$

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 $a$ pick some $k$ with $s (k) = a$ ") and it is avoided outright, because $\min$ is canonical: every nonempty set of naturals has a least element (The well-ordering principle), so $j (a)$ is determined by $a$ and $s$ alone.

Facts & Assumptions

Given: A nonempty set $A$ . For $a \in A$ and a function $s : N \to A$ write $s^{- 1} ({a}) = {k \in N : s (k) = a}$ .

[L1]

$A$ is at most countable when $A \approx n$ for some $n \in N$ or $A \approx N$ ; $A \approx 0 = \emptyset$ holds only for $A = \emptyset$ (Finite, countably infinite, countable, uncountable, The natural numbers $N$ (von Neumann)).

[L2]

Bijections, injections, surjections, images and the symmetry and transitivity of $\approx$ ; an injection is a bijection onto its image (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L3]

Well-ordering: every nonempty subset of $N$ has a least element (The well-ordering principle).

[L4]

Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L5]

For naturals, $k \in n ⟺ k < n$ , so a natural number $n$ is the set ${k \in N : k < n}$ of naturals below it; in particular $0 \in n$ whenever $n \neq 0$ (On $N$ the order is membership: $m < n ⟺ m \in n$ , proved earlier on this page from the additive order of Order on the natural numbers on the von Neumann naturals of The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1assume-hypgivenL1L2

For the forward implication assume $A$ is at most countable; since $A \neq \emptyset$ we have $A \approx N$ , or $A \approx n$ for some $n \in N$ with $n \neq 0$ , and in either case fix a bijection $ψ$ from $N$ , respectively from $n$ , onto $A$ .

1.2assume-hyp

For the converse implication assume a surjection $s : N \to A$ is given.

2.1step 1.1L2L5construct

If $ψ$ is defined on $N$ it is itself a surjection $N \to A$ ; if $ψ$ is defined on $n \neq 0$ , then $0 \in n$ by [L5] and the function $s_{ψ} : N \to A$ with $s_{ψ} (k) = ψ (k)$ for $k < n$ and $s_{ψ} (k) = ψ (0)$ for $k \geq n$ is a surjection, since every element of $A$ is $ψ (k)$ for some $k \in n$ . In both cases a surjection $N \to A$ exists.

2.2step 1.2L3construct

For each $a \in A$ the set $s^{- 1} ({a})$ is a nonempty subset of $N$ , because $s$ is surjective, so [L3] provides its least element and $j (a) : = \min s^{- 1} ({a})$ defines a function $j : A \to N$ ; no selection is made, since the least element is uniquely determined.

3.1step 2.2L2

$j$ is injective: if $j (a) = j (a^{'}) = k$ then $s (k) = a$ and $s (k) = a^{'}$ , because $j (a) \in s^{- 1} ({a})$ and $j (a^{'}) \in s^{- 1} ({a^{'}})$ , so $a = a^{'}$ .

4.1step 3.1L1L2L4

Hence $j$ is a bijection of $A$ onto $j [A] \subseteq N$ , so $A \approx j [A]$ ; the subset $j [A]$ of the at most countable set $N$ is at most countable by [L4], and transitivity of $\approx$ transfers this to $A$ .

5.1step 2.1step 3.1step 4.1∎

The forward implication is step 2.1 and the converse is step 4.1, so for nonempty $A$ countability and the existence of a surjection $N \to A$ are equivalent, with $j$ of step 2.2 the promised injection $A \to N$ .

Remarks

The hypothesis $A \neq \emptyset$ cannot be dropped in the forward direction: $\emptyset$ is finite, hence at most countable, but no function $N \to \emptyset$ exists at all. The converse direction needs no such hypothesis, since a surjection onto $A$ already forces $A \neq \emptyset$ .
Combining the two directions: a nonempty $A$ is at most countable if and only if $A ⪯ N$ (Equinumerous sets, $A \approx B$ and $A ⪯ B$ ). The forward direction of that reformulation is immediate, and the backward direction is step 4.1.
The lemma is what licenses the informal phrase "enumerate $A$ as $a_{0}, a_{1}, a_{2}, \dots$ , 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 ${A C}_{ω}$ repeat.

A nonempty set is at most countable iff it is a surjective image of $N$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A nonempty set is at most countable iff it is a surjective image of NN

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A nonempty set is at most countable iff it is a surjective image of $N$