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For limit ordinals $\alpha \le \beta$, the structures $V_{\alpha }$ and $V_{\beta }$ agree on whether any shared set is a cardinal number. They agree on being an ordinal. If $\delta <\alpha$ is not a cardinal number, a bijection from some $\gamma <\delta$ onto $\delta$ has rank of a set at most $\delta +3$, hence belongs to $V_{\alpha }$. Its defining properties are bounded formulas in set theory. Both structures therefore see the same failure witness. This full agreement is stronger than downward absoluteness of cardinalhood between arbitrary transitive models.

For a forcing atom $p$ in a ground-model forcing order, $G_p$ is a ground-model generic filter. It contains $p$ and is upward closed. For $q,r\in G_p$, choose $q^{\prime }\le p,q$ and $r^{\prime }\le p,r$. The atom property gives $s\le q^{\prime },r^{\prime }$, so $s\in G_p$ is a common strengthening of $q,r$. Thus $G_p$ is a filter in an ordered set. Every dense subset of a forcing order has some $d\le p$, and $d\in G_p$. The defining compatibility predicate is bounded to the ground-model set of conditions, so axiom schema of separation forms $G_p$ inside the ground model. For a nonminimal atom, this filter can strictly contain the principal filter above $p$.

An order is atomless if it has no forcing atom, equivalently every condition has a pair of incompatible forcing conditions below it. For a fixed order belonging to a transitive model, this property is absolute because its quantifiers range over the unchanged set of conditions. Infinite-domain Fn forcing with at least two possible values is atomless: prescribe different values at one fresh coordinate.

Every set belongs to some level of the constructible hierarchy. The assertion is $\forall x\exists \alpha (x\in L_{\alpha })$, not that each set is definable without parameters. The constructible universe theorem supplies a relative-consistency model of this assertion from ZFC.

The axiom of countable choice allows one to choose an element from each member of a countable family of nonempty sets. It is a restricted form of the axiom of choice. In the usual proof that a countable union of countable sets is countable, it suffices to choose one enumerating injection for each set; if those enumerations are already supplied, that proof requires no further choice.

A binary operation on a set $S$ is a function $S\times S\to S$.

A Borel set of Lebesgue measure zero. Every Lebesgue-null set is contained in a Borel null set, by outer regularity; consequently testing all Borel null sets suffices in the definition of a Sierpiński set.

For an infinite almost disjoint family on omega of size less than the bounding number, select countably many members and remove their intersections with earlier selected members to obtain infinite disjoint sets $C_n$. Bound, eventually and simultaneously, the finite intersections of each member with $C_n$. Choose one point of each $C_n$ above its bound. The resulting infinite set is almost disjoint from every original member, proving the family is not maximal. A selected member has one exceptional infinite intersection, which is ignored in its bounding function.

The cardinality $|X|$ of a set $X$ is the cardinal number shared by all sets in bijection with $X$. For a finite set, it is the number of elements.

The category $\mathbf{Part}$ has sets as objects and partial functions as morphisms. Composition is defined where both successive functions are defined. The nowhere-defined map is a zero morphism, and the empty set is its sole actual zero object. Adjoining a tagged basepoint turns a partial function into a total basepoint-preserving function, giving an equivalence of categories with the category of pointed sets.

The category of sets has sets as objects and functions as morphisms.

If the ground model has axiom of choice, well-order the set of subnames appearing in a forcing name for an arbitrary extension set. Evaluate those activated by the generic filter. Each element of the extension set has a least preimage index; ordering by these indices well-orders the set. Only ground choice and the ZF part of the forcing theorem are needed.

If a forcing is $\kappa$-closed in the ground model, then it adds no functions from a ground ordinal $\alpha <\kappa$ into a ground set $B$. Recursively decide each value inside the ground model and take common stronger bounds at limit stages and after the final step. Conditions deciding a whole ground function are dense below any condition asserting this type of function. The dense-below generic meeting lemma ensures the generic filter meets that dense subset of a forcing order. One arbitrary bound need not be in the filter in an ordered set.

For a limit ordinal $\kappa$, a subset $C\subseteq \kappa$ is club if it is unbounded and contains each limit below $\kappa$ of an increasing sequence from $C$. At an uncountable regular cardinal $\kappa$, intersections of fewer than $\kappa$ club sets are club: above a starting point cycle through the sets and take suprema, using regularity to stay below $\kappa$; closure puts the resulting limit in all of them. Club sets provide the reflecting ranks in worldly cardinals below an inaccessible cardinal.

At a nonzero limit ordinal $\delta$, a strictly increasing continuous sequence of cardinals $({\kappa }_{\alpha }{)}_{\alpha \le \delta }$ satisfies $\mathrm{cf}({\kappa }_{\delta })=\mathrm{cf}(\delta )$. A cofinal sequence of indices gives the upper bound. Conversely, bounds for a cofinal set of values must have cofinal indices, giving the lower bound. In particular the assertion applies to Beth numbers.