proof of is equivalent to and continuum hypothesis
The proof that implies both and that for every , are given in the entries for and .
Since there are only bounded subsets of , there is a surjective function where is the bounded subsets of . Define a sequence by if and otherwise. Since the set of such that is unbounded for any bounded subset , it follow that every bounded subset of occurs times in .
We can define a new sequence, such that for some . We can show that satisfies .
First, for any , means that for some , and since , we have .
Next take any . We consider two cases:
The set of such that forms an unbounded sequence , so there is a stationary such that . For each such , for some . But each such is equal to , so .
We define a function as follows:
To find , take . This is a bounded subset of , so is equal to an unbounded series of elements of . Take , where is the least number greater than any element of such that .
Let . This is obviously unbounded, and so there is a stationary such that .
Next, consider , the set of ordinals less than closed under . Clearly it is unbounded, since if then includes for , and so induction gives an ordinal greater than closed under (essentially the result of applying an infinite number of times). Also, is closed: take any and suppose . Then for any , there is some such that and therefore . So is closed under , and therefore contained in .
Since is a club, is stationary. Suppose . Then where . Since , , and therefore . Next take any . Since , it is closed under , hence there is some such that . Since , there is some such that , so . Since , , and since , for any , and in particular . Since we showed above that , we have for any .
|Title||proof of is equivalent to and continuum hypothesis|
|Date of creation||2013-03-22 12:53:57|
|Last modified on||2013-03-22 12:53:57|
|Last modified by||Henry (455)|
|Synonym||proof that diamond is equivalent to club and continuum hypothesis|