Timeline for Universally measurable sets and the perfect set property
Current License: CC BY-SA 3.0
8 events
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Apr 3, 2012 at 8:15 | comment | added | Detelin | Apparently, should have been more precise when stating the question. I am assuming ZFC. For the question I have asked, a set is having PSP if it is either countable or contains a perfect subset. So, apparently in ZFC the answer is NO since, as Clinton mentioned, there are universally null sets. Thanks. | |
Mar 28, 2012 at 16:33 | comment | added | Clinton Conley | Well as mentioned above, it a theorem of ZFC that the answer is "no, since there's an uncountable universally null set." If Detelin is not assuming at least ZFC, some clarification is in order. | |
Mar 28, 2012 at 13:57 | comment | added | Ed Dean | Perhaps the question's assertion that projective sets are all universally measurable and have the PSP means that Detelin wants to assume projective determinacy. | |
Mar 28, 2012 at 13:52 | comment | added | Clinton Conley | I assumed perfect set property meant the standard thing, although in retrospect it's somewhat suspicious that projective sets are claimed to possess it. (Also I wish we could edit comments -- I was so thrown by "nullness" vs. "nullity" that I forgot to demote "universally" from adverb to adjective.) | |
Mar 28, 2012 at 13:31 | comment | added | Gerald Edgar | I guess also require $\mu$ to be atomless. | |
Mar 28, 2012 at 13:29 | comment | added | Gerald Edgar | Perhaps the "perfect set property" is: given any nonzero finite measure $\mu$ on the sigma-algebra, there is a perfect set $F$ with $\mu(F)>0$. | |
Mar 28, 2012 at 13:14 | comment | added | Clinton Conley | Are you working in ZFC? If so, you can take any uncountable universally null set as a counterexample. If it contained a continuous injective image of $2^\omega$, you'd be able to push forward the product measure to contradict its universally nullness (nullity?). | |
Mar 28, 2012 at 12:43 | history | asked | Detelin | CC BY-SA 3.0 |