$exists^infty$-elimination and model companion.











up vote
0
down vote

favorite
1












In the book Ziegler, Tent: A course in model theory it states




Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




I define the theory



$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










share|cite|improve this question


























    up vote
    0
    down vote

    favorite
    1












    In the book Ziegler, Tent: A course in model theory it states




    Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
    languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




    I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




    Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
    $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
    elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




    I define the theory



    $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
    elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



    To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1





      In the book Ziegler, Tent: A course in model theory it states




      Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
      languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




      I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




      Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
      $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




      I define the theory



      $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



      To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










      share|cite|improve this question













      In the book Ziegler, Tent: A course in model theory it states




      Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
      languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




      I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




      Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
      $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




      I define the theory



      $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



      To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?







      model-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      Zikrunumea

      816




      816



























          active

          oldest

          votes











          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',
          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%2f2997862%2fexists-infty-elimination-and-model-companion%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown






























          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997862%2fexists-infty-elimination-and-model-companion%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

          Plaza Victoria

          Puebla de Zaragoza

          Musa