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Joel David Hamkins

mathematics and philosophy of the infinite

Joel David Hamkins

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Tag Archives: Woodin cardinals

Small forcing creates neither strong nor Woodin cardinals

Posted on September 25, 2011 by Joel David Hamkins
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[bibtex key=HamkinsWoodin2000:SmallForcing]

After small forcing, almost every strongness embedding is the lift of a strongness embedding in the ground model. Consequently, small forcing creates neither strong nor Woodin cardinals.

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Posted in Publications | Tagged forcing, indestructibility, W. Hugh Woodin, Woodin cardinals | Leave a reply

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Proof and the Art of Mathematics, MIT Press, 2020

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  • Comment by Joel David Hamkins on Quantifier Elimination - basic but confusing (for me) questions
    Ah, you are right (and I have now edited). What is the abstract model-theoretic characterization of QE along these lines? Or does one always have to do the hands-on inductive argument?
  • Comment by Joel David Hamkins on Quantifier Elimination - basic but confusing (for me) questions
    I also have voted to close, in light of what Emil found.
  • Answer by Joel David Hamkins for Quantifier Elimination - basic but confusing (for me) questions
    It is fairly easy to see that the second theory does not admit elimination of quantifiers. Consider the assertion "the equivalence class of x has size two." This is expressible with quantifiers as ∃y∀z(z≡x⟹z=x∨z=y). But I claim it cannot be equivalent to a quantifier free assertion, since such assertions […]
  • Comment by Joel David Hamkins on What is the relationship between "translation" and time complexity?
    Thanks for this info.
  • Comment by Joel David Hamkins on References for incompleteness proofs using infinite trees or König's lemma
    Sorry, no. But you could send the link to this post, at least, for this argument. Perhaps there is another argument...If you find out, please post it as another answer.
  • Answer by Joel David Hamkins for Model of ZFC In Which Martin's Cone Theorem Fails?
    If you have a projective well-ordering of the reals in order type ω1, as you do in L, then you can simply build one by transfinite recursion. Well order the reals, which gives you also a well-ordering of the Turing degrees. There are continuum many cones, and so we can build our desired set in […]
  • Answer by Joel David Hamkins for Computably constructing a set that is not in a free ultrafilter
    The answer is negative. For some ultrafilters U and processes S:f↦S(f), there is no total computable increasing function f with S(f)∉U. Let me explain how to make this situation occur. What I claim is that we can produce a uniform computable process f↦S(f), defined in terms of the programs, such that for […]
  • Comment by Joel David Hamkins on Computably constructing a set that is not in a free ultrafilter
    @BPP The nonexistence of such an f occurs precisely when U contains every relevant S(f), and this can happen only when the S(f) have the FIP for all total increasing f. Furthermore, when that FIP situation happens, then there is an ultrafilter containing all the S(f). So a negative answer to your question occurs exactly […]

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