Axiom of dependent choice

Inmathematics,theaxiom of dependent choice,denoted by${\mathsf {DC}}$,is a weak form of theaxiom of choice(${\mathsf {AC}}$) that is still sufficient to develop much ofreal analysis.It was introduced byPaul Bernaysin a 1942 article thatexploreswhichset-theoreticaxiomsare needed to develop analysis.

Formal statement

Ahomogeneous relation$R$on$X$is called atotal relationif for every$a\in X,$there exists some$b\in X$such that$a\,R~b$is true.

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

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

In fact,x₀may be taken to be any desired element ofX.(To see this, apply the axiom as stated above to the set of finite sequences that start withx₀and in which subsequent terms are in relation$R$,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$X$above is restricted to be the set of allreal numbers,then the resulting axiom is denoted by${\mathsf {DC}}_{\mathbb {R} }.$

Use

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

The axiom${\mathsf {DC}}$is the fragment of${\mathsf {AC}}$that is required to show the existence of a sequence constructed bytransfinite recursionofcountablelength, 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${\mathsf {ZF}}$(Zermelo–Fraenkel set theorywithout the axiom of choice),${\mathsf {DC}}$is equivalent to theBaire category theoremfor complete metric spaces.

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

${\mathsf {DC}}$is also equivalent over${\mathsf {ZF}}$to the statement that everypruned treewith$\ Omega$levels has abranch(proof below).

Furthermore,${\mathsf {DC}}$is equivalent to a weakened form ofZorn's lemma;specifically${\mathsf {DC}}$is equivalent to the statement that anypartial ordersuch that everywell-orderedchainis finite and bounded, must have a maximal element.

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\!:$$\langle x_{0},\dots,x_{n},x_{n+1}\rangle \in T,$which implies$x_{n}R\,x_{n+1}.$Therefore,${\mathsf {DC}}$is true. $(\,\Longrightarrow \,)$Let$T$be a pruned tree on$X$with$\ Omega$levels. The strategy is to define a binary relation$R$on$T$so that${\mathsf {DC}}$produces a sequence$t_{n}=\langle x_{0},\dots,x_{f(n)}\rangle$where$t_{n}R\,t_{n+1}$and$f(n)$is astrictly increasingfunction. Then the infinite sequence$\langle x_{0},\dots,x_{k},\dots \rangle$is a branch. (This proof only needs to prove this for$f(n)=m+n.$) Start by defining$u\,R\,v$if$u$is an initial subsequence of$v,\operatorname {length} (u)>0,\,$and$\operatorname {length} (v)=$$\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 anon-measurableset of real numbers, or that there is a set of real numbers without theproperty of Baireor without theperfect set property.This follows because theSolovay modelsatisfies${\mathsf {ZF}}+{\mathsf {DC}}$,and every set of real numbers in this model isLebesgue measurable,has the Baire property and has the perfect set property.

The axiom of dependent choice implies theaxiom of countable choiceand is strictly stronger.

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.

References

• Jech, Thomas (2003).Set Theory(Third Millennium ed.).Springer-Verlag.ISBN3-540-44085-2.OCLC174929965.Zbl1007.03002.