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Given finite lists ${\ell }_0,{\ell }_1,\dots$ in a set $X$, their union of entries has an explicit enumeration from a subset of ${\mathbb{N}}^2$: use the existing positions in each list and a Cantor pairing function. If the union is infinite, successively take the first occurrence of an entry not previously taken, in the order of its natural-number code. The well-order on natural numbers makes each step uniquely determined and yields an injection $\mathbb{N}\to X$. No axiom of choice is needed. This does not assert that an arbitrary countable family of unordered finite sets has a countable union in ZF without the axiom of choice.

Use “countable” to include finite and empty sets: a countable set admits an injection into ${\mathbb{N}}_0$. If $B\subseteq A$ and $j:A\to {\mathbb{N}}_0$ is injective, restricting $j$ to $B$ proves that $B$ is a countable set.

For a countable union of countable sets, index the family as $A_0,A_1,\dots$ and choose an injection $j_i:A_i\to {\mathbb{N}}_0$ for each member. For an element $a$ of the union, let $i(a)$ be the least index containing it, and assignThis is injective: equality of the pair gives the same set index and then the same element. Finally the Cantor pairing functioninjects ${\mathbb{N}}_0^2$ into ${\mathbb{N}}_0$. The union is therefore countable. The proof uses chosen injections for the family, as in the usual set-theoretic framework with axiom of countable choice.

Choose injections $a:X\to \mathbb{N}$ and $b:Y\to \mathbb{N}$. The map $(x,y)\mapsto (a(x),b(y))$ is an injection into the countable set $\mathbb{N}\times \mathbb{N}$. Composing with the Cantor pairing function gives an injection into $\mathbb{N}$. Thus the Cartesian product of two countable sets is countable, including the case of an empty factor.

Choose an injection $j:A\to \mathbb{N}$. For each $b\in B$, its fiber under the surjection $f$ is nonempty, so defineTwo distinct fibers cannot contain the same label $j(a)$, because $j$ is an injection and $f(a)$ has a single value. Thus $h:B\to \mathbb{N}$ is an injection, proving that $B$ is a countable set. Taking the least label avoids any need for a separate choice of one element from each fiber. If $A$ is empty, surjectivity makes $B$ empty as well.

For the unheaded enumeration request, list pairs of positive natural numbers by their sum. An explicit Cantor pairing function isFor $d=m+n-2\ge 0$, the positive integers $n=1,\dots ,d+1$ give the consecutive labels $d(d+1)/2+1,\dots ,(d+1)(d+2)/2$. These blocks are disjoint, consecutive and cover $\mathbb{N}$. Consequently $\pi$ is a bijection from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$, so $\mathbb{N}\times \mathbb{N}$ is countable.

Work in ZF with the law of excluded middle, without the axiom of choice. The paper's terminology is an infinite Dedekind-finite set. The useful common principle is countable union of explicitly ordered finite lists without choice: given an actual sequence of finite lists, enumerate their entries by list number and position using a Cantor pairing function. If the union of entries is infinite, repeatedly take the entry with the least code not already selected. This produces an injection from $\mathbb{N}$ without choosing any enumerations of unordered sets.

Consequently, if a family of objects has an explicitly ordered finite list of labels for each object, and only finitely many objects can be made from any given finite pool of labels, then a countably infinite list of distinct objects forces a countably infinite subset of the label set. For a Dedekind-finite label set this is impossible. If the label set embeds into the object family as well, infinitude is preserved. ThusThe distinction between ordered lists supplied by the data and arbitrary finite subsets is essential: a blanket countable-union theorem for unordered finite sets would introduce a choice principle.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$; equivalently, its elements can be listed in a sequence. The integers are countable, for example in the order $0,1,-1,2,-2,\dots$. Enumerating pairs by diagonals, or applying the Cantor pairing function, then shows that $\mathbb{N}\times \mathbb{Z}$ is countable.

Suppose instead that the power set $\mathcal{P}(\mathbb{N})$ had an enumeration $A_1,A_2,\dots$. The Cantor diagonal argument formsFor every $n$, the set $D$ differs from $A_n$ at the element $n$, so it is absent from the alleged enumeration. Therefore

Write the sets as $A_1,A_2,\dots$, and list each nonempty countable set as $A_i=\{a_{i1},a_{i2},\dots \}$. Enumerate the pairs $(i,j)\in {\mathbb{N}}^2$ along successive diagonals, as in the Cantor pairing function, and output $a_{ij}$ when it has not appeared before. Every member of ${\bigcup }_i{A}_i$ eventually appears, proving that a countable union of countable sets is countable.

List pairs $(m,n)\in {\mathbb{N}}^2$ by increasing value of $m+n$, and within each diagonal by increasing $m$:Every pair occurs after finitely many earlier diagonals. More explicitly, the Cantor pairing functionis an injection ${\mathbb{N}}^2\to \mathbb{N}$. Hence $\mathbb{N}\times \mathbb{N}$ is countable.