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May 19, 2013 at 14:54 vote accept Jing Zhang
May 19, 2013 at 14:24 comment added François G. Dorais I don't think different approaches to forcing disagree in this case and none of them add new natural numbers. You can use the computability theoretic approach if you want to. In that case the only actual name in the forcing language is the name for the generic function $g$. Then $(s,j) \Vdash g(x) = y$ holds iff $x < |s|$ and $s(x) = y$ and the rest follows the usual inductive definition of forcing. Note that some things are different from Cohen forcing, for example $(s,j) \Vdash g(x) \geq y$ iff $\varphi_j(x) \geq y$ since there is no extension of $(s,j)$ that forces $g(x) = z$ where $z < y$.
May 19, 2013 at 8:20 comment added Jing Zhang What is the forcing language here (sorry for getting into messy details)? For example, what is $(s,j)\Vdash \bar{n}\in X$? In the normal context of strings, $\sigma\Vdash \bar{n}\in X \leftrightarrow \sigma(n)=1$ (I got from Odifreddi's book). In addition, my intention was to preserve the first-order universe so that $B\Sigma_2^0$ is still false. But I am not sure whether the forcing mentioned here would alter the first order universe. Thanks!
May 18, 2013 at 11:37 comment added François G. Dorais By $g \geq f$, I mean $g(x) \geq f(x)$ for all $x$. I never assumed that the universe was standard, so this works as is in non-standard universes. In particular, this shows that the domination principle does not imply $\Sigma^0_2$-bounding.
May 18, 2013 at 10:21 comment added Jing Zhang @François: Thanks! Can I just have it clarified what it means to say $g\geq f$ in the definition of poset? Since I am interested in the non-standard universe, do you think the same argument goes through (it occurs to me so).
May 17, 2013 at 19:07 history edited François G. Dorais CC BY-SA 3.0
s/RCA/ACA/ near end
May 17, 2013 at 17:38 history edited François G. Dorais CC BY-SA 3.0
fixed grammar
May 17, 2013 at 17:23 history answered François G. Dorais CC BY-SA 3.0