Ultrafilters as a double dualIs the non-triviality of the algebraic dual of an infinite-dimensional vector...



Ultrafilters as a double dual


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8












$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:





  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.


These are reminiscent of similar claims that can be made about vector spaces and double duals:





  • $V$ canonically embeds into $V^{star star}$;

  • If $V$ is finite-dimensional, then we have $V = V^{star star}$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.


This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?




  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.


Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.






ct.category-theory lo.logic gn.general-topology ultrafilters







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble♦
    3 hours ago








  • 2




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    2 hours ago












  • $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    2 hours ago
















8












$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:





  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.


These are reminiscent of similar claims that can be made about vector spaces and double duals:





  • $V$ canonically embeds into $V^{star star}$;

  • If $V$ is finite-dimensional, then we have $V = V^{star star}$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.


This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?




  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.


Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.






ct.category-theory lo.logic gn.general-topology ultrafilters







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble♦
    3 hours ago








  • 2




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    2 hours ago












  • $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    2 hours ago














8












8








8





$begingroup$


Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:





  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.


These are reminiscent of similar claims that can be made about vector spaces and double duals:





  • $V$ canonically embeds into $V^{star star}$;

  • If $V$ is finite-dimensional, then we have $V = V^{star star}$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.


This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?




  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.


Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.






ct.category-theory lo.logic gn.general-topology ultrafilters







share|cite|improve this question











$endgroup$




Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:





  • $X$ canonically embeds into $beta X$ (by taking principal ultrafilters);

  • If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.

  • If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.


These are reminiscent of similar claims that can be made about vector spaces and double duals:





  • $V$ canonically embeds into $V^{star star}$;

  • If $V$ is finite-dimensional, then we have $V = V^{star star}$;

  • If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.


This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?




  • The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;

  • If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);

  • If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.


Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.






ct.category-theory lo.logic gn.general-topology ultrafilters




ct.category-theory lo.logic gn.general-topology ultrafilters






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









YCor

28.1k483136




28.1k483136










asked 3 hours ago









Adam P. GoucherAdam P. Goucher

6,53322856




6,53322856








  • 12




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble♦
    3 hours ago








  • 2




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    2 hours ago












  • $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    2 hours ago














  • 12




    $begingroup$
    Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
    $endgroup$
    – Todd Trimble♦
    3 hours ago








  • 2




    $begingroup$
    Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
    $endgroup$
    – YCor
    2 hours ago












  • $begingroup$
    Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
    $endgroup$
    – Asaf Karagila
    2 hours ago








12




12




$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
3 hours ago






$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
3 hours ago






2




2




$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
2 hours ago






$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
2 hours ago














$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
2 hours ago




$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
2 hours ago










1 Answer
1






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$begingroup$

This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.






share|cite|improve this answer









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    1 Answer
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    active

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    5












    $begingroup$

    This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



    Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



      Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



        Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.






        share|cite|improve this answer









        $endgroup$



        This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.



        Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Nik WeaverNik Weaver

        21.3k148130




        21.3k148130






























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