Axiom of dependent choice

Use

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 ${\displaystyle \mathsf{D}\mathsf{C}}$ is the fragment of ${\displaystyle \mathsf{A}\mathsf{C}}$ 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 ${\displaystyle \mathsf{Z}\mathsf{F}}$ (Zermelo–Fraenkel set theory without the axiom of choice), ${\displaystyle \mathsf{D}\mathsf{C}}$ is equivalent to the Baire category theorem for complete metric spaces.

It is also equivalent over ${\displaystyle \mathsf{Z}\mathsf{F}}$ to the downward Löwenheim–Skolem theorem.

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

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

Relation with other axioms

Unlike full ${\displaystyle \mathsf{A}\mathsf{C}}$, ${\displaystyle \mathsf{D}\mathsf{C}}$ is insufficient to prove (given ${\displaystyle \mathsf{Z}\mathsf{F}}$) 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 ${\displaystyle \mathsf{Z}\mathsf{F}+\mathsf{D}\mathsf{C}}$, 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.

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.