Prove that the topological closure of a set is definable if the set is definable












1












$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18
















1












$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18














1












1








1





$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$




Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory




logic first-order-logic predicate-logic model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 19:57









Asaf Karagila♦

305k33436766




305k33436766










asked Dec 5 '18 at 19:11









quanticboltquanticbolt

769512




769512








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18














  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18








1




1




$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54




$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54












$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04




$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04












$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12






$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12














$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15






$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15






1




1




$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18




$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18










1 Answer
1






active

oldest

votes


















1












$begingroup$

As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






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: "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%2f3027507%2fprove-that-the-topological-closure-of-a-set-is-definable-if-the-set-is-definable%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









    1












    $begingroup$

    As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



    $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



    You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



      $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



      You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



        $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



        You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






        share|cite|improve this answer









        $endgroup$



        As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



        $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



        You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 1:42









        Alex KruckmanAlex Kruckman

        27.6k32658




        27.6k32658






























            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%2f3027507%2fprove-that-the-topological-closure-of-a-set-is-definable-if-the-set-is-definable%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more