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A subset $E$ of a set $X$ is cofinite in $X$ when its complement $X\setminus E$ is a finite set. For infinite $X$, the cofinite subsets form the cofinite filter. A nonprincipal ultrafilter on $\mathbb{N}$ contains every cofinite subset: it contains no singleton, hence no finite set, and decides between each set and its complement.

If $M$ is a transitive model of ZF, its internally computed $L_{\alpha }$ is the actual $L_{\alpha }$ for each ordinal $\alpha \in M$. At a successor stage use absoluteness of satisfaction for the same set structure, formula codes and parameters; at a limit stage take the union of the already identical earlier stages. Choice is not required.

If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.

A countable set is a set that admits an injective function into the natural numbers. It is therefore either a finite set or a countably infinite set.

For measurable sets $E_n$, a measure satisfiesReplace $E_n$ by the disjoint sets $E_n\setminus {\bigcup }_{j<n}{E}_j$, use countable additivity, and then monotonicity. The same property is an axiom for an outer measure on arbitrary sets.

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.

The cubic lattice in dimension $d$ has graph vertex set ${\mathbb{Z}}^d$ and an edge between $x,y$ exactly when ${\sum }_i|x_i-y_i|=1$. For $d=2$ it is the square lattice. Every graph vertex has $2d$ graph neighbours, so it is a locally finite graph.

A set $X$ is Dedekind-finite when no injective function $\mathbb{N}\to X$ exists. In classical Zermelo–Fraenkel set theory this is equivalent to $X$ not being in bijection with a proper subset: an injective non-surjective self-map generates distinct iterates from an element outside its image, and a countably infinite subset permits a shift fixing its complement. A finite set is Dedekind-finite. Without the axiom of choice, some infinite sets can also be Dedekind-finite.

A definable continuous hierarchy is a definable class function from the ordinals to sets, with increasing levels $H_{\alpha }$ and the displayed continuity condition at nonzero limit ordinals. Its union is a definable class in set theory. One often additionally requires transitive sets as levels. Finite collections of first-order formulas reflect along this hierarchy by the reflection theorem for definable hierarchies.

If $p\in G$ and a ground-model set $D$ is dense below a forcing condition $p$, then the generic filter $G$ meets $D$. Add all incompatible forcing conditions with $p$ to $D$. The enlarged set is globally dense: a condition compatible with $p$ first has a common strengthening, then a further strengthening in $D$. Genericity meets the enlargement, and directedness excludes its incompatible part.

A finite open coordinate set witnesses an increasing event when prescribing those coordinates open forces the event, regardless of the remaining configuration. The disjoint occurrence of increasing events $A\square B$ means that $A$ and $B$ have disjoint finite witnesses. For finite-coordinate events this is the usual disjoint-occurrence definition. The van den Berg-Kesten inequality bounds its probability by $\mathbb{P}(A)\mathbb{P}(B)$ under an independent Bernoulli distribution product measure.

A product over the empty set has value one, the identity for multiplication. This makes splitting a product into successive pieces valid even when one piece contains no factors. In particular, the factorial convention is $0!=1$.

The equivalence class of an element $x$ under an equivalence relation $R$ is the set of all elements related to $x$. Two such sets are equal or disjoint, and together they partition the original set.