In mathematics, the axiom of dependent choice, denoted by ${\displaystyle {\mathsf {DC}}}$, is a weak form of the axiom of choice (${\displaystyle {\mathsf {AC}}}$) that is still sufficient to develop much of real analysis. It was introduced by Paul Bernays in a 1942 article in reverse mathematics that explores which set-theoretic axioms are needed to develop analysis.^[a]

A homogeneous relation ${\displaystyle R}$ on ${\displaystyle X}$ is called a total relation if for every ${\displaystyle a\in X,}$ there exists some ${\displaystyle b\in X}$ such that ${\displaystyle a\,R~b}$ is true.

The axiom of dependent choice can be stated as follows: For every nonempty set ${\displaystyle X}$ and every total relation ${\displaystyle R}$ on ${\displaystyle X,}$ there exists a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle X}$ such that

${\displaystyle x_{n}\,R~x_{n+1}}$ for all ${\displaystyle n\in \mathbb {N} .}$

In fact, x₀ may be taken to be any desired element of X. (To see this, apply the axiom as stated above to the set of finite sequences that start with x₀ and in which subsequent terms are in relation ${\displaystyle R}$, together with the total relation on this set of the second sequence being obtained from the first by appending a single term.)

If the set ${\displaystyle X}$ above is restricted to be the set of all real numbers, then the resulting axiom is denoted by ${\displaystyle {\mathsf {DC}}_{\mathbb {R} }.}$

Even without such an axiom, for any ${\displaystyle n}$, one can use ordinary mathematical induction to form the first ${\displaystyle n}$ terms of such a sequence. The axiom of dependent choice says that we can form a whole (countably infinite) sequence this way.

The axiom ${\displaystyle {\mathsf {DC}}}$ is the fragment of ${\displaystyle {\mathsf {AC}}}$ that is required to show the existence of a sequence constructed by transfinite recursion of countable length, if it is necessary to make a choice at each step and if some of those choices cannot be made independently of previous choices.

Over ${\displaystyle {\mathsf {ZF}}}$ (Zermelo–Fraenkel set theory without the axiom of choice), ${\displaystyle {\mathsf {DC}}}$ is equivalent to the Baire category theorem for complete metric spaces.^[1]

It is also equivalent over ${\displaystyle {\mathsf {ZF}}}$ to the downward Löwenheim–Skolem theorem.^[b][2]

${\displaystyle {\mathsf {DC}}}$ is also equivalent over ${\displaystyle {\mathsf {ZF}}}$ to the statement that every pruned tree with ${\displaystyle \omega }$ levels has a branch (proof below).

Furthermore, ${\displaystyle {\mathsf {DC}}}$ is equivalent to a weakened form of Zorn's lemma; specifically ${\displaystyle {\mathsf {DC}}}$ is equivalent to the statement that any partial order such that every well-ordered chain is finite and bounded, must have a maximal element.^[3]

Proof that ${\displaystyle \,{\mathsf {DC}}\iff }$ Every pruned tree with ω levels has a branch

${\displaystyle (\,\Longleftarrow \,)}$ Let ${\displaystyle R}$ be an entire binary relation on ${\displaystyle X}$. The strategy is to define a tree ${\displaystyle T}$ on ${\displaystyle X}$ of finite sequences whose neighboring elements satisfy ${\displaystyle R.}$ Then a branch through ${\displaystyle T}$ is an infinite sequence whose neighboring elements satisfy ${\displaystyle R.}$ Start by defining ${\displaystyle \langle x_{0},\dots ,x_{n}\rangle \in T}$ if ${\displaystyle x_{k}R\,x_{k+1}}$ for ${\displaystyle 0\leq k<n.}$ Since ${\displaystyle R}$ is entire, ${\displaystyle T}$ is a pruned tree with ${\displaystyle \omega }$ levels. Thus, ${\displaystyle T}$ has a branch ${\displaystyle \langle x_{0},\dots ,x_{n},\dots \rangle .}$ So, for all ${\displaystyle n\geq 0\!:}$ ${\displaystyle \langle x_{0},\dots ,x_{n},x_{n+1}\rangle \in T,}$ which implies ${\displaystyle x_{n}R\,x_{n+1}.}$ Therefore, ${\displaystyle {\mathsf {DC}}}$ is true. ${\displaystyle (\,\Longrightarrow \,)}$ Let ${\displaystyle T}$ be a pruned tree on ${\displaystyle X}$ with ${\displaystyle \omega }$ levels. The strategy is to define a binary relation ${\displaystyle R}$ on ${\displaystyle T}$ so that ${\displaystyle {\mathsf {DC}}}$ produces a sequence ${\displaystyle t_{n}=\langle x_{0},\dots ,x_{f(n)}\rangle }$ where ${\displaystyle t_{n}R\,t_{n+1}}$ and ${\displaystyle f(n)}$ is a strictly increasing function. Then the infinite sequence ${\displaystyle \langle x_{0},\dots ,x_{k},\dots \rangle }$ is a branch. (This proof only needs to prove this for ${\displaystyle f(n)=m+n.}$) Start by defining ${\displaystyle u\,R\,v}$ if ${\displaystyle u}$ is an initial subsequence of ${\displaystyle v,\operatorname {length} (u)>0,\,}$ and ${\displaystyle \operatorname {length} (v)=}$ ${\displaystyle \operatorname {length} (u)+1.}$ Since ${\displaystyle T}$ is a pruned tree with ${\displaystyle \omega }$ levels, ${\displaystyle R}$ is entire. Therefore, ${\displaystyle {\mathsf {DC}}}$ implies that there is an infinite sequence ${\displaystyle t_{n}}$ such that ${\displaystyle t_{n}\,R\,t_{n+1}.}$ Now ${\displaystyle t_{0}=\langle x_{0},\dots ,x_{m}\rangle }$ for some ${\displaystyle m\geq 0.}$ Let ${\displaystyle x_{m+n}}$ be the last element of ${\displaystyle t_{n}.}$ Then ${\displaystyle t_{n}=\langle x_{0},\dots ,x_{m},\dots ,x_{m+n}\rangle .}$ For all ${\displaystyle k\geq 0,}$ the sequence ${\displaystyle \langle x_{0},\dots ,x_{k}\rangle }$ belongs to ${\displaystyle T}$ because it is an initial subsequence of ${\displaystyle t_{0}\,(k\leq m)}$ or it is a ${\displaystyle t_{n}\,(k\geq m).}$ Therefore, ${\displaystyle \langle x_{0},\dots ,x_{k},\dots \rangle }$ is a branch.

Unlike full ${\displaystyle {\mathsf {AC}}}$, ${\displaystyle {\mathsf {DC}}}$ is insufficient to prove (given ${\displaystyle {\mathsf {ZF}}}$) that there is a non-measurable set of real numbers, or that there is a set of real numbers without the property of Baire or without the perfect set property. This follows because the Solovay model satisfies ${\displaystyle {\mathsf {ZF}}+{\mathsf {DC}}}$, and every set of real numbers in this model is Lebesgue measurable, has the Baire property and has the perfect set property.

The axiom of dependent choice implies the axiom of countable choice and is strictly stronger.^[4][5]

It is possible to generalize the axiom to produce transfinite sequences. If these are allowed to be arbitrarily long, then it becomes equivalent to the full axiom of choice.

• Jech, Thomas (2003). Set Theory (Third Millennium ed.). Springer-Verlag. ISBN 3-540-44085-2. OCLC 174929965. Zbl 1007.03002.