Parts of Set Theory immune to independence - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:16:31Z http://mathoverflow.net/feeds/question/58276 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58276/parts-of-set-theory-immune-to-independence Parts of Set Theory immune to independence Tanmay Inamdar 2011-03-12T18:13:03Z 2011-10-29T06:57:56Z <p>The motivation for asking this question is a passage (3.2) in an <a href="http://books.google.co.in/books?id=adfidZ9Ojg8C&amp;lpg=PP1&amp;dq=in%20scope%20of%20logic&amp;pg=PA108#v=onepage&amp;q&amp;f=false" rel="nofollow">article</a> by Greg Hjorth where he said that "...it is also an attractive feature of the theory of Borel cardinalities and of the theory of $L(\mathbb{R})$ cardinality that these are largely immunized against independence." </p> <p>My question is: What other parts of Set Theory are so immunised? Assuming something extra is okay(say Determinacy), but whatever it is, should settle 'most' of the questions.</p> <p>The wording above is not too clear, but I'm not sure how I can make a stronger statement(suggestions about this would be great). I guess that if you ask high-level enough questions(definability hierarchy wise), independence will come in sooner or later. However, I'm not really asking this from the point of view of absoluteness, but from the point of view of what large class of questions can be settled by what small set of tools.</p> <p>A bonus question is: Why is the part about Borel cardinalities true? I guess this might have some simple absoluteness explanation, but the 'largely' tells me there is more to it than I think.</p> <p>Thanks in advance.</p> <p>P.S. I'm not sure if I've tagged this appropriately, perhaps a 'soft-question' or 'big-list' one would be okay (although I wonder how big the list would be!). I hope people will retag it if they find it suitable.</p> http://mathoverflow.net/questions/58276/parts-of-set-theory-immune-to-independence/58285#58285 Answer by Justin Moore for Parts of Set Theory immune to independence Justin Moore 2011-03-12T20:48:38Z 2011-05-25T16:09:58Z <p>It is a theorem of Woodin that if there is a proper class of Woodin cardinals, then the theory of $L(\mathbb{R})$ can not be changed by forcing. Since forcing and large cardinals are essentially our only means for establishing independence results, this can be interpreted as saying that the theory of $L(\mathbb{R})$ is immune to independence phenomena (except for that which G\" odel's theorem imposes). Here $L(\mathbb{R})$ is the smallest model of ZF which contains all of the reals. $L(\mathbb{R})$ does satisfy the Axiom of Dependent Choice under this assumption, as well as the Axiom of Determinacy. Most of not all theorems in real and complex analysis, measure theory (in the setting of standard Borel space), manifolds, geometry, and number theory can be regarded as statements about what is true in $L(\mathbb{R})$. It is the ideal model in which to study descriptive set theory. Uncountable sets and cardinals, however, often behave strangely in this model (largely because of the influence of the Axiom of Determinacy). For instance, in $L(\mathbb{R})$ and under the above assumptions, $\omega_1$ and $\omega_2$ are measurable cardinals, $\omega_n$ is a singular cardinal for each $n > 2$, there is no uncountable well orderable set of reals, and there are no non-principal ultrafilters on $\omega$. </p> <p>Ironically, Woodin's theorem was the culmination of several deep results concerning iterated forcing, the combinatorics of $\omega_1$, the study of large cardinals, and the fine structure of inner models generalizing $L$.</p> http://mathoverflow.net/questions/58276/parts-of-set-theory-immune-to-independence/65977#65977 Answer by Timothy Chow for Parts of Set Theory immune to independence Timothy Chow 2011-05-25T17:31:51Z 2011-05-25T17:31:51Z <p>I'm not sure if this is the kind of thing you're looking for, but the assumption $V=L$ settles most "interesting" questions in set theory. For example, it implies the generalized continuum hypothesis (and therefore the axiom of choice) and it disproves the existence of measurable cardinals.</p> <p>However, most (though not all) set theorists don't believe that $V$ is "really" equal to $L$, so the "immunity to independence" enjoyed by $V=L$ doesn't excite people as much as you might naively expect.</p> http://mathoverflow.net/questions/58276/parts-of-set-theory-immune-to-independence/79453#79453 Answer by Trevor Wilson for Parts of Set Theory immune to independence Trevor Wilson 2011-10-29T06:57:56Z 2011-10-29T06:57:56Z <p>Some of Shelah's work has shown that parts of set theory that seemed to be dominated by independence results (for example, cardinal arithmetic) actually have a lot of nontrivial structure that is provable in ZFC. I don't know enough about the area to say any more, but I can point to Shelah's survey paper "Cardinal arithmetic for skeptics."</p>