User m_korch - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:31:32Z http://mathoverflow.net/feeds/user/31944 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123693/random-reals-and-strongly-meager-sets Random reals and strongly meager sets m_korch 2013-03-06T02:14:15Z 2013-03-06T06:34:21Z <p>Adding a single Cohen real makes the set of reals from the ground model strong measure zero (see <a href="http://mathoverflow.net/questions/63497/cohen-reals-and-strong-measure-zero-sets" rel="nofollow">this question</a>).</p> <p>The notion of strong measure zero sets has its dual concept in the category branch -- strongly meager sets. A set $X\subseteq \mathbb{R}$ is strongly meager if for any null set $Y$ there exists $t\in \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. One can see duality of these notions due to Galvin-Mycielski-Solovay Theorem which states that a set $X\subseteq \mathbb{R}$ is strong measure zero if and only if for any meager set $Y$ there exists $t\in \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. </p> <p>Random real forcing is dual to Cohen forcing in the sense of measure and category. Therefore it makes sense to ask, whether:</p> <blockquote> <p>The set of reals from generic model $\mathbb{R}\cap V$ is strongly meager after adding a single random real?</p> </blockquote> <p>I have heard that the answer is affirmative, but I have not been able to find any published proof. Note that $\mathbb{R}\cap V$ is meager after adding a random real (see <a href="http://mathoverflow.net/questions/8269/adding-a-random-real-makes-the-set-of-ground-model-reals-meager" rel="nofollow">this question</a>).</p>