1. Formal statement

2. Use

3. Equivalent statements

4. Relation with other axioms

5. Notes

6. References

In mathematics, the axiom of dependent choice, denoted by , is a weak form of the axiom of choice () 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.^[1]

Formal statement

The axiom can be stated as follows: For every nonempty set and every entire binary relation on there exists a sequence in such that for all (Here, an entire binary relation on is one where for every there exists some such that is true.) Even without such an axiom, one can use ordinary mathematical induction to form the first terms of such a sequence, for every the axiom of dependent choice says that we can form a whole sequence this way.

If the set above is restricted to be the set of all real numbers, then the resulting axiom is denoted by

Use

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.

Equivalent statements

Over Zermelo–Fraenkel set theory , is equivalent to the Baire category theorem for complete metric spaces.^[2]

It is also equivalent over to the Löwenheim–Skolem theorem.^[3][4]

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

Proof that Every pruned tree with ω levels has a branch

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.

Relation with other axioms

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.^[5][6]

Notes

References

1. ^"The foundation of analysis does not require the full generality of set theory but can be accomplished within a more restricted frame." {{cite journal |last=Bernays |first=Paul |year=1942 |series=A system of axiomatic set theory |title=Part III. Infinity and enumerability. Analysis. |journal=Journal of Symbolic Logic |volume=7 |page=65 |mr=0006333 |doi=10.2307/2266303 |jstor=2266303}} The axiom of dependent choice is stated on p. 86.
2. ^"The Baire category theorem implies the principle of dependent choices." {{cite journal |author=Blair, Charles E. |year=1977 |title=The Baire category theorem implies the principle of dependent choices |journal=Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. |volume=25 |issue=10 |pages=933–934}}
3. ^Moore states that "Principle of Dependent Choices Löwenheim–Skolem theorem" — that is, implies the Löwenheim–Skolem theorem. See table {{cite book |last=Moore |first=Gregory H. |year=1982 |title=Zermelo's Axiom of Choice: Its origins, development, and influence |page=325 |publisher=Springer |isbn=0-387-90670-3}}
4. ^The converse is proved in {{cite book |last1=Boolos |first1=George S. |author1link=George Boolos |last2=Jeffrey |first2=Richard C. |author2link=Richard Jeffrey |year=1989 |title=Computability and Logic |edition=3rd |pages=155–156 |publisher=Cambridge University Press |isbn=0-521-38026-X}}
5. ^Bernays proved that the axiom of dependent choice implies the axiom of countable choice See esp. p. 86 in {{cite journal |last=Bernays |first=Paul |year=1942 |series=A system of axiomatic set theory |title=Part III. Infinity and enumerability. Analysis. |journal=Journal of Symbolic Logic |volume=7 |pages=65–89 |mr=0006333 |doi=10.2307/2266303 |jstor=2266303}}
6. ^For a proof that the Axiom of Countable Choice does not imply the Axiom of Dependent Choice see {{Citation |last=Jech |first=Thomas |authorlink=Thomas Jech |year=1973 |title=The Axiom of Choice |pages=130–131 |publisher=North Holland |isbn=978-0-486-46624-8}}

• {{cite book |last=Jech |first=Thomas |title=Set Theory |year=2003 |edition=Third Millennium |publisher=Springer-Verlag |isbn=3-540-44085-2 |oclc=174929965 |zbl=1007.03002}}

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