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Is this question acceptable?: Request for re-open

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    • CommentAuthoryfyf
    • CommentTimeMar 5th 2011
     
    Hi meta-MO,

    I am requesting a re-open of my question here:
    http://mathoverflow.net/questions/57426/are-there-any-non-linear-solutions-of-cauchys-equation-fxyfxfy-with

    I think it isn't a bad question, I was just slightly ambiguous with the formulation and it would be nice to have it answered just for future reference for others.

    Thanks!
  2.  
    You are asking if it is possible to prove in ZF that nonlinear solutions to the equation exist. But as George Lowther points out in his comment there is a model of ZF where there are no nonlinear solutions.
  3.  

    Now that the question has been changed, I totally agree the question has been answered by George. (If existence were provable in ZF, then it would be provable in ZF + AD, and then existence would hold in any model of ZF + AD. But it doesn't, as George explained.) My original interpretation of the question was this: is there some weakened form of the axiom of choice, call it WAC, so that existence of nonlinear solutions can be proven in ZF + WAC?

    There is no doubt an uninteresting sense in which the answer is yes, since we are only dealing with functions on R, whereas AC is a much more global statement about general sets.

    • CommentAuthoryfyf
    • CommentTimeMar 5th 2011
     
    Yes, I agree that George answered the question in that sense, but why does it have to be closed instead of receiving a proper answer (not in comment form) and getting marked as answered, at least for archival purposes?
  5.  

    @yfyf: I don't know, maybe you're right, although it seems very strange to me to reopen an edited question which was answered even before it had been properly asked!

    The custom for many MO participants is that if a question can be answered easily in a brief comment, then it will be. (People may be somewhat demure about being rewarded for easy and well-known answers.) There may also be a perception that not a lot of thought went into the question, because it was somewhat confusingly formulated in the first place.

    Since the question hasn't vanished into the ether just by being closed, a more satisfactory way to proceed may be to ask a better (and more carefully thought-out) question, and link back to the question which was closed as having provided the initial impetus.

  6.  
    It is well known that there are no measurable solutions (this is even an exercise in several books). It is well-known that "all functions are measurable" is consistent with ZF (this is a famous result of Solovay, we do not even need to invoke determinacy here). This solves the original question. It seems to me that, as originally stated, the question was not appropriate for MathOverflow. This is not to say in the least there are no interesting mathematics attached to it.

    EDIT: Now that I see your edited version, I agree that there are some interesting mathematics there (specifically, whether ZF suffices, since appeal to Solovay's result requires the consistency of inaccessible cardinals, and AD requires much more). I have voted to Re-open.
  7.  

    we do not even need to invoke determinacy here

    No, but it's one avenue; Solovay's model on the other hand does require an inaccessible cardinal (according to wikipedia; I'm not a set theorist). I guess most set theorists believe that's a much less drastic assumption than determinacy. Either way, yes, the question was solved.

  8.  
    @Todd: "Either way, yes, the question was solved." Not the re-edited version.

    Let me be more precise: The consistency strength of determinacy is much much higher than that of an inaccessible. In fact, it is precisely that of infinitely many Woodin cardinals. If k is a Woodin cardinal, there are k strongly inaccessible cardinals below k. (But this is a very meek statement. It is like saying that I am shorter than the Everest.)

    Solovay's result *does* require an inaccessible, this was shown by Shelah. So the remaining question (part of the re-edited version) is whether the inaccessible is still needed just to guarantee the existence of nonlinear functions, which in principle is weaker than measurability of all functions.
  9.  

    So the remaining question (part of the re-edited version) is whether the inaccessible is still needed just to guarantee the existence of nonlinear functions, which in principle is weaker than measurability of all functions.

    No. Automatic continuity also holds for Baire measurable homomorphisms from R to R. Shelah showed that one can get ZF + DC + "All sets of reals have the Baire property" without an inaccessible.

  10.  
    @François: Yes, I know. But I think this deserves to be explicitly answered in MO rather than meta, and there was the suggestion above that if the question was already answered in comments, then it didn't need to be reopened, so I didn't want to just point this out here, and kill the question's chances of surviving.
    • CommentAuthorJDH
    • CommentTimeMar 5th 2011
     

    I agree that these answers should be made explicitly on MO. I have voted to reopen.

    There is another aspect to the question: the existence of such a function is evidently a weak choice principle---how does it relate to the other choice principles?

  12.  

    @Andres: I should have mentioned that I agree that the question should be reopened. I'll be the fifth vote...

    @JDH: The statement is Form 366 in Consequences of the Axiom of Choice. The web interface http://consequences.emich.edu/CONSEQ.HTM didn't give anything useful. I don't have a copy of the book handy right now.

  13.  
    @François: I just checked. They do not mention any implications other than he existence of non-linear functions implies the existence of non-measurable sets. The only model for its negation that is mentioned is Solovay's (in particular, it may be worthwhile to email them and ask them to include mention of Shelah's result).
  14.  

    Andres wrote: @Todd: "Either way, yes, the question was solved." Not the re-edited version.

    I want to make sure I understand this. The re-edited question reads:

    Can existence of non-linear solutions be proven from ZF alone?

    I had thought the answer was "no". For if ZF could prove existence of non-linear solutions, then non-linear solutions exist in any model of ZF, and in particular in Solovay's model.

    Is the issue then that Solovay's model might not "exist", i.e., that ZF + (exists inaccessible) might be inconsistent?

    "So the remaining question (part of the re-edited version) is whether the inaccessible is still needed just to guarantee the existence of nonlinear functions, which in principle is weaker than measurability of all functions."

    I guess you meant to say the non-existence of nonlinear functions?

    @JDH: "There is another aspect to the question: the existence of such a function is evidently a weak choice principle---how does it relate to the other choice principles?" I also mentioned this as my interpretation of the original problem, but AFAICT the re-edited question doesn't ask this.

  15.  
    Todd: What you say is essentially correct. For sure, inaccessible cardinals are consistent (Joel will disagree, but they are). But assuming only the consistency of ZF we cannot build Solovay's model.

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