Filters on a set of filters, are they equivalent to just filters?












2












$begingroup$


Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.



Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?






set-theory filters







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
    $endgroup$
    – Patrick Da Silva
    Jul 19 '14 at 18:35










  • $begingroup$
    @PatrickDaSilva $U$ is a set. It does not "have poset structure".
    $endgroup$
    – porton
    Jul 19 '14 at 18:39










  • $begingroup$
    Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 7:00












  • $begingroup$
    Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 8:02


















2












$begingroup$


Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.



Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?






set-theory filters







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
    $endgroup$
    – Patrick Da Silva
    Jul 19 '14 at 18:35










  • $begingroup$
    @PatrickDaSilva $U$ is a set. It does not "have poset structure".
    $endgroup$
    – porton
    Jul 19 '14 at 18:39










  • $begingroup$
    Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 7:00












  • $begingroup$
    Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 8:02
















2












2








2





$begingroup$


Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.



Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?






set-theory filters







share|cite|improve this question











$endgroup$




Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.



Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?






set-theory filters




set-theory filters






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 19 '14 at 18:16







porton

















asked Jul 19 '14 at 18:03









portonporton

1,92611229




1,92611229












  • $begingroup$
    Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
    $endgroup$
    – Patrick Da Silva
    Jul 19 '14 at 18:35










  • $begingroup$
    @PatrickDaSilva $U$ is a set. It does not "have poset structure".
    $endgroup$
    – porton
    Jul 19 '14 at 18:39










  • $begingroup$
    Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 7:00












  • $begingroup$
    Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 8:02




















  • $begingroup$
    Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
    $endgroup$
    – Patrick Da Silva
    Jul 19 '14 at 18:35










  • $begingroup$
    @PatrickDaSilva $U$ is a set. It does not "have poset structure".
    $endgroup$
    – porton
    Jul 19 '14 at 18:39










  • $begingroup$
    Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 7:00












  • $begingroup$
    Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
    $endgroup$
    – Patrick Da Silva
    Jul 20 '14 at 8:02


















$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35




$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35












$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39




$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39












$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00






$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00














$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02






$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02












1 Answer
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oldest

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$begingroup$

No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).



On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.



Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.






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    $begingroup$

    No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).



    On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.



    Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).



      On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.



      Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).



        On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.



        Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.






        share|cite|improve this answer









        $endgroup$



        No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).



        On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.



        Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 6:26









        Eric WofseyEric Wofsey

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