A set is said to be embeddable in another set if there is a one-to-one function . For example, every subset of a set is embeddable in the set. In particular, is embeddable in every set. Clearly, given any set, there is an ordinal embeddable in it. On the other hand, is any set embeddable in an ordinal? If so, then the set is well-orderable (as it is equipollent to one), which is equivalent to the well-ordering principle. In other words, in ZF, AC is equivalent to saying that every set is embeddable in an ordinal. Without AC, how much can one deduce? In 1915, Hartogs proved the following:
Given any set , there is an ordinal not embeddable in .
Let be the class of all ordinals embeddable in . We want to show that is in fact an ordinal, not embeddable in . We have the following steps:
is a set.
Let be the subset of , the powerset of , consisting of all well-orderable subsets of . For each element of , let be the collection of well-orderings on . Each element of is a subset of , so that . For any , the well-ordered set is order isomorphic to exactly one element of . Conversely, every , by definition, is embeddable in , so equipollent to a subset of . We may well-order via , and this well-ordering . Therefore, there is a surjection from onto . Since is a set, so is its range (by the replacement axiom), which is just .
is an ordinal.
is not embeddable in .
Otherwise, , contradicting the fact that an ordinal can never be a member of itself.
The proof is done within ZF, without the aid of the axiom of choice.
In fact, the constructed above is the Hartogs number of , for if is another ordinal distinct from that is not embeddable in , then , so by trichotomy.
Remark. For every set , its Hartogs number is a cardinal number: it is first of all an ordinal, so , where is the ordering on the ordinals, and if (meaning ), then is embeddable in . Since is equipollent to , is embeddable in , contradicting the definition of . Hence . From the discussion so far, we see that can be thought of as a class function from the class of all sets onto the class Cn of all cardinal numbers. In addition, assuming AC, every set is well-orderable, so that is the least cardinal greater , for every set .
|Date of creation||2013-03-22 18:49:48|
|Last modified on||2013-03-22 18:49:48|
|Last modified by||CWoo (3771)|