- #1
KOSS
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Can any math geeks help?
Refering to the talk,
(Woodin: plenary talk at the 2010 International Congress of Mathematicians) http://bitcast-a.bitgravity.com/highbrow/livearchive40009/26aug-13.45to14.45.flv
In Hugh Woodin's 2010 ICM talk on Ultimate L he introduces Godel's constructible universe of sets L then discusses the possibility that the universe of sets V is exactly L and then brings up Scott's theorem which says that if V=L then there are no measurable cardinals.
Then Woodin says something like, "so V=L is denied by infinity." (About 28 minutes into the video talk linked above.)
Q. What does he mean by this?
My problem understanding this statement is that he seems to be saying that large measurable cardinals must somehow be desirable in set theory. Why? Or if I interpret the remark as meaning that any sort of infinite set is denied by V=L then I'm just not groking this, because I always though Godel's L does admit infinite sets, just not these large cardinals.
So is Woodin really only paraphrasing something like "so V=L is denied by those who would assume large cardinals are useful," or is he stating something stronger or more subtle?
The best i can make of Woodin's remark is that he means Scott's theorem implies that if V=L then "there are no interesting large cardinals." In which case my question would be what the heck does "an interesting large cardinal" mean?
PLEASE: no comments necessary from finitists or physicists or anyone else who does not believe in either the consistency or utility of transfinite numbers. I respect your beliefs without needing to agree.
Refering to the talk,
(Woodin: plenary talk at the 2010 International Congress of Mathematicians) http://bitcast-a.bitgravity.com/highbrow/livearchive40009/26aug-13.45to14.45.flv
In Hugh Woodin's 2010 ICM talk on Ultimate L he introduces Godel's constructible universe of sets L then discusses the possibility that the universe of sets V is exactly L and then brings up Scott's theorem which says that if V=L then there are no measurable cardinals.
Then Woodin says something like, "so V=L is denied by infinity." (About 28 minutes into the video talk linked above.)
Q. What does he mean by this?
My problem understanding this statement is that he seems to be saying that large measurable cardinals must somehow be desirable in set theory. Why? Or if I interpret the remark as meaning that any sort of infinite set is denied by V=L then I'm just not groking this, because I always though Godel's L does admit infinite sets, just not these large cardinals.
So is Woodin really only paraphrasing something like "so V=L is denied by those who would assume large cardinals are useful," or is he stating something stronger or more subtle?
The best i can make of Woodin's remark is that he means Scott's theorem implies that if V=L then "there are no interesting large cardinals." In which case my question would be what the heck does "an interesting large cardinal" mean?
PLEASE: no comments necessary from finitists or physicists or anyone else who does not believe in either the consistency or utility of transfinite numbers. I respect your beliefs without needing to agree.