A question on the ultrafilter number












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Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.



Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?






set-theory lo.logic







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    2












    $begingroup$


    Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.



    Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?






    set-theory lo.logic







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.



      Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?






      set-theory lo.logic







      share|cite|improve this question









      $endgroup$




      Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.



      Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?






      set-theory lo.logic




      set-theory lo.logic






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      asked 5 hours ago









      IsaacIsaac

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      384






















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          $begingroup$

          The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:



          Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
          Ultrafilters with small generating sets.
          Israel J. Math. 65 (1989), no. 3, 259–271.






          share|cite|improve this answer









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          • $begingroup$
            Thanks Andreas.
            $endgroup$
            – Isaac
            3 hours ago











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          1 Answer
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          active

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          4












          $begingroup$

          The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:



          Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
          Ultrafilters with small generating sets.
          Israel J. Math. 65 (1989), no. 3, 259–271.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks Andreas.
            $endgroup$
            – Isaac
            3 hours ago
















          4












          $begingroup$

          The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:



          Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
          Ultrafilters with small generating sets.
          Israel J. Math. 65 (1989), no. 3, 259–271.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks Andreas.
            $endgroup$
            – Isaac
            3 hours ago














          4












          4








          4





          $begingroup$

          The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:



          Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
          Ultrafilters with small generating sets.
          Israel J. Math. 65 (1989), no. 3, 259–271.






          share|cite|improve this answer









          $endgroup$



          The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:



          Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
          Ultrafilters with small generating sets.
          Israel J. Math. 65 (1989), no. 3, 259–271.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Andreas BlassAndreas Blass

          58k7138224




          58k7138224












          • $begingroup$
            Thanks Andreas.
            $endgroup$
            – Isaac
            3 hours ago


















          • $begingroup$
            Thanks Andreas.
            $endgroup$
            – Isaac
            3 hours ago
















          $begingroup$
          Thanks Andreas.
          $endgroup$
          – Isaac
          3 hours ago




          $begingroup$
          Thanks Andreas.
          $endgroup$
          – Isaac
          3 hours ago


















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