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equivalence of Zorn's lemma and the axiom of choice
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(Proof)
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Let be a set partially ordered by such that each chain has an upper bound. Define
. Let
. If
then it follows that is maximal.
Suppose no
. Then by the axiom of choice there is a choice function on , and since for each we have
, it follows that . Define
for all ordinals by transfinite induction:


And for a limit ordinal , let
be an upper bound of for .
This construction can go on forever, for any ordinal. Then we can easily construct an injective function from to by
for an arbitrary . This must be injective, since
implies
. But that requires that be a proper class, in contradiction to the fact that it is a set. So there can be no such choice function, and there must be a maximal element of .
For the reverse, assume Zorn's lemma and let be any set of non-empty sets. Consider the set of functions
partially ordered by inclusion. Then the union of any chain in is also a member of (since the union of a chain of functions is always a function). By Zorn's lemma, has a maximal element , and since any
function with domain smaller than can be easily expanded,
, and so is a choice function for .
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"equivalence of Zorn's lemma and the axiom of choice" is owned by Henry.
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(view preamble)
Cross-references: expanded, domain, union, inclusion, functions, Zorn's lemma, maximal element, contradiction, proper class, implies, injective, injective function, limit ordinal, transfinite induction, ordinals, choice function, upper bound, chain
There are 2 references to this entry.
This is version 7 of equivalence of Zorn's lemma and the axiom of choice, born on 2002-08-25, modified 2005-11-27.
Object id is 3358, canonical name is ProofOfZornsLemma.
Accessed 5665 times total.
Classification:
| AMS MSC: | 03E25 (Mathematical logic and foundations :: Set theory :: Axiom of choice and related propositions) |
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Pending Errata and Addenda
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