Do complete atomic boolean algebras form a full subcategory of boolean algebras?












3












$begingroup$


I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?



I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.



Thank you!






boolean-algebra duality-theorems







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?



    I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.



    Thank you!






    boolean-algebra duality-theorems







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?



      I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.



      Thank you!






      boolean-algebra duality-theorems







      share|cite|improve this question









      $endgroup$




      I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?



      I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.



      Thank you!






      boolean-algebra duality-theorems




      boolean-algebra duality-theorems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 28 at 14:18









      L. De RudderL. De Rudder

      335




      335






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.

            So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.



            Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.

            Let $Y$ be the set of ultrafilters of $mathbf A$;
            for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).

            Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.



            Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
            $$Phi(a) = { U in Y : a in U }$$
            is a Boolean algebra homomorphism.

            Now consider the set $At_A$ of atoms of $mathbf A$.
            As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.

            For $a = {x}$, with $x in X$ (that is, $a in At_A$),
            $$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
            whence $Phi(At_A) subseteq At_B$.

            If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.

            But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
            $$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
            where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).



            We conclude that $Phi$ is not a complete homomorphism.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
              $endgroup$
              – Eric Wofsey
              Jan 29 at 21:05










            • $begingroup$
              @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
              $endgroup$
              – amrsa
              Jan 29 at 22:43












            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%2f3090909%2fdo-complete-atomic-boolean-algebras-form-a-full-subcategory-of-boolean-algebras%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.






                share|cite|improve this answer









                $endgroup$



                No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 29 at 21:04









                Eric WofseyEric Wofsey

                191k14216349




                191k14216349























                    2












                    $begingroup$

                    Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.

                    So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.



                    Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.

                    Let $Y$ be the set of ultrafilters of $mathbf A$;
                    for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).

                    Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.



                    Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
                    $$Phi(a) = { U in Y : a in U }$$
                    is a Boolean algebra homomorphism.

                    Now consider the set $At_A$ of atoms of $mathbf A$.
                    As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.

                    For $a = {x}$, with $x in X$ (that is, $a in At_A$),
                    $$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
                    whence $Phi(At_A) subseteq At_B$.

                    If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.

                    But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
                    $$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
                    where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).



                    We conclude that $Phi$ is not a complete homomorphism.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                      $endgroup$
                      – Eric Wofsey
                      Jan 29 at 21:05










                    • $begingroup$
                      @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                      $endgroup$
                      – amrsa
                      Jan 29 at 22:43
















                    2












                    $begingroup$

                    Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.

                    So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.



                    Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.

                    Let $Y$ be the set of ultrafilters of $mathbf A$;
                    for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).

                    Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.



                    Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
                    $$Phi(a) = { U in Y : a in U }$$
                    is a Boolean algebra homomorphism.

                    Now consider the set $At_A$ of atoms of $mathbf A$.
                    As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.

                    For $a = {x}$, with $x in X$ (that is, $a in At_A$),
                    $$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
                    whence $Phi(At_A) subseteq At_B$.

                    If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.

                    But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
                    $$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
                    where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).



                    We conclude that $Phi$ is not a complete homomorphism.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                      $endgroup$
                      – Eric Wofsey
                      Jan 29 at 21:05










                    • $begingroup$
                      @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                      $endgroup$
                      – amrsa
                      Jan 29 at 22:43














                    2












                    2








                    2





                    $begingroup$

                    Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.

                    So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.



                    Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.

                    Let $Y$ be the set of ultrafilters of $mathbf A$;
                    for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).

                    Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.



                    Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
                    $$Phi(a) = { U in Y : a in U }$$
                    is a Boolean algebra homomorphism.

                    Now consider the set $At_A$ of atoms of $mathbf A$.
                    As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.

                    For $a = {x}$, with $x in X$ (that is, $a in At_A$),
                    $$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
                    whence $Phi(At_A) subseteq At_B$.

                    If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.

                    But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
                    $$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
                    where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).



                    We conclude that $Phi$ is not a complete homomorphism.






                    share|cite|improve this answer











                    $endgroup$



                    Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.

                    So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.



                    Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.

                    Let $Y$ be the set of ultrafilters of $mathbf A$;
                    for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).

                    Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.



                    Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
                    $$Phi(a) = { U in Y : a in U }$$
                    is a Boolean algebra homomorphism.

                    Now consider the set $At_A$ of atoms of $mathbf A$.
                    As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.

                    For $a = {x}$, with $x in X$ (that is, $a in At_A$),
                    $$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
                    whence $Phi(At_A) subseteq At_B$.

                    If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.

                    But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
                    $$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
                    where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).



                    We conclude that $Phi$ is not a complete homomorphism.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 29 at 22:46

























                    answered Jan 29 at 18:36









                    amrsaamrsa

                    3,8552618




                    3,8552618








                    • 1




                      $begingroup$
                      $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                      $endgroup$
                      – Eric Wofsey
                      Jan 29 at 21:05










                    • $begingroup$
                      @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                      $endgroup$
                      – amrsa
                      Jan 29 at 22:43














                    • 1




                      $begingroup$
                      $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                      $endgroup$
                      – Eric Wofsey
                      Jan 29 at 21:05










                    • $begingroup$
                      @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                      $endgroup$
                      – amrsa
                      Jan 29 at 22:43








                    1




                    1




                    $begingroup$
                    $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                    $endgroup$
                    – Eric Wofsey
                    Jan 29 at 21:05




                    $begingroup$
                    $bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
                    $endgroup$
                    – Eric Wofsey
                    Jan 29 at 21:05












                    $begingroup$
                    @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                    $endgroup$
                    – amrsa
                    Jan 29 at 22:43




                    $begingroup$
                    @EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
                    $endgroup$
                    – amrsa
                    Jan 29 at 22:43


















                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090909%2fdo-complete-atomic-boolean-algebras-form-a-full-subcategory-of-boolean-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

                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                    Image
                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                    Read more

                    Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                    Image
                    up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                    Read more

                    A Topological Invariant for $pi_3(U(n))$

                    Image
                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                    Read more