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Possible Duplicate:
Examples of ZFC theorems proved via forcing

In Partitions of partial orders , we see two proofs of a corollary. One of them is an argument without forcing while the other one uses a method of forcing to justify the existence of objects in the ground model having certain properties. Philosophically, this is different than forcing to produce new models of set theory to show relative consistency results (e.g. $\lnot$CH, the failure of the continuum hypothesis with ZFC).

My question is:

Are there results that have relatively simple proofs via forcing arguments of the first kind but seem to become much more technical when carried out without the use of forcing?

As a related example, if we have a supercompact cardinal, then we can force to an extension $V[G]$ where PFA (the proper forcing axiom), which asserts the existence of $\mathcal{D}$-generic filters for every $\omega_1$ collection of dense subsets $\mathcal{D}$ from a proper partial order, is true. To do this, we fix a proper partial order $\mathbb{Q} \in V[G]$ and lift a ground model $\theta$-supercompactness embedding $j_{\mathbb{Q}}: V \rightarrow M$ to $j_{\mathbb{Q}}: V[G] \rightarrow M[j(G)]$ in $V[j(G)]$ for some sufficiently large $\theta$. In $V[j(G)]$, we then argue that we've performed enough forcing for $V[G]$ to have a filter meeting any collection of $\omega_1$ many dense subsets from $\mathbb{Q}$.

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  • $\begingroup$ Related: mathoverflow.net/questions/29945/… and mathoverflow.net/questions/42569/… . $\endgroup$ Dec 9, 2010 at 0:58
  • $\begingroup$ Hi Jason, I believe your question is essentially a duplicate: mathoverflow.net/questions/42569/… I have two answers there. In both, there are results that we do not know how to prove without forcing; although one of them is an "internal" forcing. The other (on Baumgartner-Hajnal) is very much in the spirit of the question you cited above. $\endgroup$ Dec 9, 2010 at 0:58
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    $\begingroup$ I asked this question mathoverflow.net/questions/47185 and the answer used forcing to make a conclusion about the ground model, In that case it was to justify the lack of an object having a certain property. $\endgroup$ Dec 9, 2010 at 2:06
  • $\begingroup$ Jason, I up-voted the question, since I think this is a fascinating topic, but meanwhile, unfortunately, the question seems to be a duplicate. $\endgroup$ Dec 9, 2010 at 3:15

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