Axiom of dependent choice

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

Formal statement

The axiom can be stated as follows: For every nonempty set $X$ and every entire binary relation $R$ on $X$, there exists a sequence $(x_{n})_{n\in \mathbb {N} }$ in $X$ such that $\ x_{n}\ R~x_{n+1}\$ for all $n\in \mathbb {N}$. (Here, an entire binary relation on $X$ is one where for every $a\in X$, there exists some $b\in X$ such that $\ a\ R~b\$ is true.) Note that even without such an axiom, one can use ordinary mathematical induction to form the first $n$ terms of such a sequence, for every $n\in \mathbb {N}$; the axiom of dependent choice says that we can form a whole sequence this way.

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

Use

${\mathsf {DC}}$ is the fragment of ${\mathsf {AC}}$ 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.

Equivalent statements

Over Zermelo–Fraenkel set theory ${\mathsf {ZF}}$, ${\mathsf {DC}}$ is equivalent to the Baire category theorem for complete metric spaces.

It is also equivalent over ${\mathsf {ZF}}$ to the Löwenheim–Skolem theorem.

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

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

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

Relation with other axioms

Unlike full ${\mathsf {AC}}$, ${\mathsf {DC}}$ is insufficient to prove (given ${\mathsf {ZF}}$) 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 ${\mathsf {ZF}}+{\mathsf {DC}}$, 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.

References

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