Probabilities independent of ZFC? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:54:00Z http://mathoverflow.net/feeds/question/56990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56990/probabilities-independent-of-zfc Probabilities independent of ZFC? sebastian 2011-03-01T11:13:00Z 2011-03-01T14:05:34Z <p>Hi guys,</p> <p>is it possible to change the probability of an event via forcing? More precisely, is there an innocent looking question on the probability of "something" whose answer is independent of ZFC?</p> <p>All the best, Sebastian</p> http://mathoverflow.net/questions/56990/probabilities-independent-of-zfc/56992#56992 Answer by Joel David Hamkins for Probabilities independent of ZFC? Joel David Hamkins 2011-03-01T11:28:12Z 2011-03-01T14:05:34Z <p>There are several issues. </p> <p>On the one hand, any set can be made countable by forcing, and this process will certainly affect the measure of the set, if it did not have measure zero in the ground model. </p> <p>But in the context of the Lebesgue measure on the reals, say, it is natural to consider not the set itself, but the Borel description of the set, interpreted first in the ground model and then reinterpreted in the forcing extension. (For exampe, the "unit interval" of $V$ is not necessarily the same as the unit interval of a forcing extension $V[G]$, but we have a borel code that correctly picks out the unit interval when interpreted in any model of ZFC.) In this case, one gets a positive solution for preservation of measure. The reason is that the assertion that the measure of the set with Borel code $b$ is $x$ has complexity at most $\Sigma^1_2(b,x)$ and hence is absolute to all forcing extensions by the Shoenfield absoluteness theorem. In this sense, the measure of a measurable set cannot be affected by forcing. </p> <p>Meanwhile, the use of other non-absolute descriptions can lead again to a negative answer, where the measure can be affected by forcing. For example, consider the set $X$ of all binary sequences $x$ whose sequence of digits is realized somewhere in the GCH pattern of cardinals, in the sense that there is an ordinal $\beta$ such that $x(n)=1$ iff $2^{\aleph_{\beta+n}}=\aleph_{\beta+n+1}$. If the Generalized Continuum Hypothesis holds, then $X$ has measure zero, since only one pattern is realized. But one can force the GCH pattern to realize all patterns, and so there are forcing extensions in which $X$ has full measure.</p> <p>Here is another comparatively concrete example. Consider the set of reals that are constructible, in the sense of <a href="http://en.wikipedia.org/wiki/Constructible_universe" rel="nofollow">G&ouml;del's constructible universe</a>. This set has complexity $\Sigma^1_2$ in the descriptive set-theoretic hierarchy, which is just a step up from Borel. The set has full measure in the constructible universe, of course, but it is easily made to have measure zero in a forcing extension. Thus, the probability that a randomly chosen real number is constructible has an answer that is independent of ZFC, because in some models of set theory this probability is 1 and in others it is 0.</p>