Suppose the least size of a non meager set of reals is $\kappa$. Is it still $\kappa$ after forcing with Random $\times$ Random?
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$\begingroup$ In the random real extension, non(Meager) is preserved. $\endgroup$– user60503Commented Oct 15, 2014 at 0:16
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1$\begingroup$ Isn't Random$\times$Random, the same as two step iteration of Random. If so, then just use your above comment two times! $\endgroup$– Mohammad GolshaniCommented Oct 15, 2014 at 3:46
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1$\begingroup$ @MohammadGolshani No. Random$\times$Random adds Cohen reals, for example. A two step iteration of Random is isomorphic to a single step. $\endgroup$– Andrés E. CaicedoCommented Oct 16, 2014 at 6:31
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$\begingroup$ I have a proof (communicated by Shelah) that non(Meager) is preserved in $V^P$ where $P$ is a finite support iteration of random forcing of length $\omega$ (so a Cohen is added in the limit). Perhaps $P$ and Random X Random are forcing equivalent? $\endgroup$– AshutoshCommented Oct 19, 2014 at 12:13
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I'm curious to hear the set-theoretic answer. In the meantime it's interesting to note that in the computability-theoretic setting the answer seems to be "no".
There the analogue would be: Suppose $A$ does not compute any weakly meager engulfing (w.m.e.) set, and $B$ is Schnorr (say) random over $A$. Then does $B\oplus A$ still not compute a w.m.e. set? But in that setting, w.m.e. is the same up to Turing degree as high-or-DNR (by link above), and each Schnorr random set does have that property by a result of Nies, Stephan, and Terwijn (see "Ref 24" in the paper linked above).
answered Oct 16, 2014 at 1:51