Representation theorem for Heyting algebras?












12














A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).



I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?



Thank you in advance.










share|cite|improve this question
























  • @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
    – Kyle
    Jul 14 '14 at 17:33










  • The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
    – Zhen Lin
    Jul 14 '14 at 17:57










  • Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
    – Nagase
    Jul 25 '14 at 5:22






  • 1




    Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
    – Mauro ALLEGRANZA
    Aug 29 '14 at 14:24


















12














A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).



I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?



Thank you in advance.










share|cite|improve this question
























  • @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
    – Kyle
    Jul 14 '14 at 17:33










  • The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
    – Zhen Lin
    Jul 14 '14 at 17:57










  • Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
    – Nagase
    Jul 25 '14 at 5:22






  • 1




    Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
    – Mauro ALLEGRANZA
    Aug 29 '14 at 14:24
















12












12








12


5





A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).



I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?



Thank you in advance.










share|cite|improve this question















A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).



I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?



Thank you in advance.







reference-request logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 14 '14 at 17:11

























asked Jul 14 '14 at 16:55









Marco Vergura

3,0811930




3,0811930












  • @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
    – Kyle
    Jul 14 '14 at 17:33










  • The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
    – Zhen Lin
    Jul 14 '14 at 17:57










  • Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
    – Nagase
    Jul 25 '14 at 5:22






  • 1




    Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
    – Mauro ALLEGRANZA
    Aug 29 '14 at 14:24




















  • @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
    – Kyle
    Jul 14 '14 at 17:33










  • The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
    – Zhen Lin
    Jul 14 '14 at 17:57










  • Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
    – Nagase
    Jul 25 '14 at 5:22






  • 1




    Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
    – Mauro ALLEGRANZA
    Aug 29 '14 at 14:24


















@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33




@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33












The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57




The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57












Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22




Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22




1




1




Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24






Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24












1 Answer
1






active

oldest

votes


















8














Esakia Duality might just be the thing that you are looking for.



For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.



This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.



Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.






share|cite|improve this answer





















    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: "69"
    };
    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%2fmath.stackexchange.com%2fquestions%2f867090%2frepresentation-theorem-for-heyting-algebras%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









    8














    Esakia Duality might just be the thing that you are looking for.



    For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
    leq,
    mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.



    This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.



    Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.






    share|cite|improve this answer


























      8














      Esakia Duality might just be the thing that you are looking for.



      For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
      leq,
      mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.



      This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.



      Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.






      share|cite|improve this answer
























        8












        8








        8






        Esakia Duality might just be the thing that you are looking for.



        For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
        leq,
        mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.



        This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.



        Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.






        share|cite|improve this answer












        Esakia Duality might just be the thing that you are looking for.



        For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
        leq,
        mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.



        This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.



        Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 14 '15 at 12:10









        FML

        18618




        18618






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • 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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2fmath.stackexchange.com%2fquestions%2f867090%2frepresentation-theorem-for-heyting-algebras%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

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅