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I am trying to understand the proof of Zorn's lemma from the axiom of choice, but I do not entirely understand the step where we create a increasing sequence (a_i) of sets in a partially ordered set S indexed by ordinals. It is defined through transfinite recursion, but how does that work?
We assume that Zorn's lemma is false for a partially ordered set S. So for any element we have a larger one. One starts with defining the function b on the set of chains of S to S by letting b(T) be an element of S such that b(T) is larger than any element of the chain. T has an upper bound, so such an element exists, and by the axiom of choice this function is well-defined. Then one picks an element a_0 in S, and then for each ordinal w>0 one defines [tex]a_w = b(\{ a_v | v<w\})[/tex] by transfinite recursion. Could someone explain this step?
The proof continues by saying that this sequence exhausts S. What does that mean? My guess would be that we can identify an ordinal W with a cardinality strictly larger than that of S (e.g. the cardinality of P(S)), so that we have an surjection from S into the set [tex]\{ a_v | v \leq W\}[/tex] which has cardinality W. But I don't know if we have ordinals corresponding to cardinals.
By the way, I would prefer that the explanation does not rely on the well-ordering theorem. This brings up the question: does the existence and properties of the ordinals needed for this proof rely on the well-ordering theorem?
We assume that Zorn's lemma is false for a partially ordered set S. So for any element we have a larger one. One starts with defining the function b on the set of chains of S to S by letting b(T) be an element of S such that b(T) is larger than any element of the chain. T has an upper bound, so such an element exists, and by the axiom of choice this function is well-defined. Then one picks an element a_0 in S, and then for each ordinal w>0 one defines [tex]a_w = b(\{ a_v | v<w\})[/tex] by transfinite recursion. Could someone explain this step?
The proof continues by saying that this sequence exhausts S. What does that mean? My guess would be that we can identify an ordinal W with a cardinality strictly larger than that of S (e.g. the cardinality of P(S)), so that we have an surjection from S into the set [tex]\{ a_v | v \leq W\}[/tex] which has cardinality W. But I don't know if we have ordinals corresponding to cardinals.
By the way, I would prefer that the explanation does not rely on the well-ordering theorem. This brings up the question: does the existence and properties of the ordinals needed for this proof rely on the well-ordering theorem?
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