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

A homogeneous relation on is called a total relation if for every there exists some such that is true.

The axiom of dependent choice can be stated as follows: For every nonempty set and every total relation on there exists a sequence in such that

- for all

In fact, *x*_{0} 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 *x*_{0} and in which subsequent terms are in relation , 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 above is restricted to be the set of all real numbers, then the resulting axiom is denoted by

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

The axiom is the fragment of 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 (Zermelo–Fraenkel set theory without the axiom of choice), is equivalent to the Baire category theorem for complete metric spaces.^{[1]}

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

is also equivalent over to the statement that every pruned tree with levels has a branch (*proof below*).

Furthermore, is equivalent to a weakened form of Zorn's lemma; specifically 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 Every pruned tree with ω levels has a branch |
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Let be an entire binary relation on . The strategy is to define a tree on of finite sequences whose neighboring elements satisfy Then a branch through is an infinite sequence whose neighboring elements satisfy Start by defining if for Since is entire, is a pruned tree with levels. Thus, has a branch So, for all which implies Therefore, is true.
Let be a pruned tree on with levels. The strategy is to define a binary relation on so that produces a sequence where and is a strictly increasing function. Then the infinite sequence is a branch. (This proof only needs to prove this for ) Start by defining if is an initial subsequence of and Since is a pruned tree with levels, is entire. Therefore, implies that there is an infinite sequence such that Now for some Let be the last element of Then For all the sequence belongs to because it is an initial subsequence of or it is a Therefore, is a branch. |

Unlike full , is insufficient to prove (given ) 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 , 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.