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



Ultrafilters as a double dual


Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?Unique Existence and the Axiom of ChoiceWhat are the algebras for the double dualization monad?Characterization of Stone-Cech compactificationsFor a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X to k$?Ultrafilters and diagonal argumentsIs there a general way to turn a 2-monad into a lax-idempotent (a.k.a. KZ) 2-monad?What, mathematically speaking, does it mean to say that the continuation monad can simulate all monads?Relation between monads, operads and algebraic theories (Again)A monad that unions sets













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






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "504"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324867%2fultrafilters-as-a-double-dual%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to MathOverflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324867%2fultrafilters-as-a-double-dual%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Szabolcs (Ungheria) Altri progetti | Menu di navigazione48°10′14.56″N 21°29′33.14″E /...

            Image
            Comuni della provincia di Szabolcs-Szatmár-Bereg Ungheria2001provincia di Szabolcs-Szatmár-Bereg Szabolcs comune Localizzazione Stato   Ungheria Regione Grande Pianura Settentrionale Provincia Szabolcs-Szatmár-Bereg Territorio Coordinate 48°10′14.56″N 21°29′33.14″E  /  48.17071°N 21.49254°E 48.17071; 21.49254  ( Szabolcs ) Coordinate: 48°10′14.56″N 21°29′33.14″E  /  48.17071°N 21.49254°E 48.17071; 21.49254  ( Szabolcs ) Superficie 5,88 km² Abitanti 425 (2001) Densità 72,28 ab./km² Altre informazioni Cod. postale 4467 Prefisso 42 Fuso orario UTC+1 Codice KSH 19169 Cartografia Szabolcs Sito istituzionale Modifica dati su Wikidata  · Manuale Szabolcs è un comune dell'Ungheria di 425 abitanti (dati 2001). È situato nella provincia di Szabolcs-Szatmár-Bereg. Altri progetti | Altri progetti Wikimedia Commons Wikimedia Commons contiene immagini o altri file su Szabolcs .mw-parser-output
            Read more

            Discografia di Klaus Schulze Indice Album in studio | Album dal vivo | Singoli | Antologie | Colonne...

            Image
            Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune
            Read more

            How to make inet_server_addr() return localhost in spite of ::1/128RETURN NEXT in Postgres FunctionConnect to...

            Image
            Potential client has a problematic employee I can't work with Can the "Friends" spell be used without making the target hostile? Does it take energy to move something in a circle? Why are carbons of Inositol chiral centers? Count repetitions of an array Which RAF squadrons and aircraft types took part in the bombing of Berlin on the 25th of August 1940? Memory usage: #define vs. static const for uint8_t Broad Strokes - missing letter riddle Why does 0.-5 evaluate to -5? Why is 'diphthong' pronounced the way it is? Why did Luke use his left hand to shoot?
            Read more