proof of complete partial orders do not add small subsets
Take any , . Let be a name for . There is some such that
For any , we construct by induction a series of elements stronger than . Each will determine whether or not . Since we know the subset is bounded below , we can use the fact that is complete to find a single element stronger than which fixes the exact value of . Since the series is definable in , so is , so we can conclude that above any element is an element which forces . Then also forces , completing the proof.
Since forcing can be described within , is a set in . Then, given any , we can define and for any (), is an element of stronger than such that either or . For limit , let be any upper bound of for (this exists since is -complete and ), and let be stronger than and satisfy either or . Finally let be the upper bound of for . since is -complete.
Note that these elements all exist since for any and any (first-order) sentence there is some such that forces either or .
not only forces that is a bounded subset of , but for every ordinal it forces whether or not that ordinal is contained in . But the set is defineable in , and is of course equal to in any generic containing . So .
Since this holds for any element stronger than , it follows that , and therefore .
|Title||proof of complete partial orders do not add small subsets|
|Date of creation||2013-03-22 12:53:35|
|Last modified on||2013-03-22 12:53:35|
|Last modified by||Henry (455)|