$mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$ is an algebra












-1












$begingroup$



Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



Show that $mathscr{C}$ is an algebra.




To show $mathscr{C}$ is an algebra:



i) $emptysetinmathscr{C}$



ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



For example if I want to check $mathscr{C}$ is closed for unions then:



If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



Question:



It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



Thanks in advance!






measure-theory proof-verification elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$



    Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



    Show that $mathscr{C}$ is an algebra.




    To show $mathscr{C}$ is an algebra:



    i) $emptysetinmathscr{C}$



    ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



    iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



    iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



    For example if I want to check $mathscr{C}$ is closed for unions then:



    If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



    Question:



    It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



    Thanks in advance!






    measure-theory proof-verification elementary-set-theory proof-writing







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$



      Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



      Show that $mathscr{C}$ is an algebra.




      To show $mathscr{C}$ is an algebra:



      i) $emptysetinmathscr{C}$



      ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



      iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



      iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



      For example if I want to check $mathscr{C}$ is closed for unions then:



      If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



      Question:



      It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



      Thanks in advance!






      measure-theory proof-verification elementary-set-theory proof-writing







      share|cite|improve this question











      $endgroup$





      Let $mathscr{R}$ be a ring of subsets of $Omega$ and $mathscr{C}={EsubsetOmega:Einmathscr{R}:text{or}:Omega-Einmathscr{R}}$.



      Show that $mathscr{C}$ is an algebra.




      To show $mathscr{C}$ is an algebra:



      i) $emptysetinmathscr{C}$



      ii) $Ainmathscr{C}$ and $Binmathscr{C}$ so $Acup Binmathscr{C}$



      iii)$Ainmathscr{C}$ and $Binmathscr{C}$ so $Asetminus Binmathscr{C}$



      iv)$Ainmathscr{C}$ so $A^c=Omegasetminus Ainmathscr{C}$



      For example if I want to check $mathscr{C}$ is closed for unions then:



      If $Ainmathscr{R}$ and $Binmathscr{R}$then $Acup Binmathscr{R}$ because $mathscr{R}$ is a ring hence closed for finite unions or $Omegasetminus(Acup B)=(Acup B)^c=A^ccap B^cinmathscr{R}$ once $mathscr{R}$ is closed for finite intersections. This concludes $Acup B$ is in $mathscr{C}$.



      Question:



      It was pointed out that this argument I wrote is wrong once I does not prove $Acup B$ is in $mathscr{C}$. But I fail to see why? Can someone tell me what is failing?



      Thanks in advance!






      measure-theory proof-verification elementary-set-theory proof-writing




      measure-theory proof-verification elementary-set-theory proof-writing






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 22:05









      Andrés E. Caicedo

      65.4k8158249




      65.4k8158249










      asked Jan 18 at 19:30









      Pedro GomesPedro Gomes

      1,8262721




      1,8262721






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15



















          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23













          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%2f3078656%2fmathscrc-e-subset-omegae-in-mathscrr-textor-omega-e-in-mathscrr%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









          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15
















          1












          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15














          1












          1








          1





          $begingroup$

          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.






          share|cite|improve this answer











          $endgroup$



          You have to take sets $A$ and $ B$ from $mathscr{C}$ and not from $mathscr{R}$. Now do some case analysis



          Case 1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          Case 2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $A-Binmathscr{R}stackrel{?}{implies} Acup Binmathscr{C}$



          Case 3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          However, I'm not sure how to deal with case 2.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 20:29

























          answered Jan 18 at 19:36









          greedoidgreedoid

          42.4k1153105




          42.4k1153105












          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15


















          • $begingroup$
            How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
            $endgroup$
            – Pedro Gomes
            Jan 18 at 19:39










          • $begingroup$
            I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Jan 18 at 21:23












          • $begingroup$
            Yes, but is it in $mathscr{C}$?
            $endgroup$
            – greedoid
            Jan 19 at 18:06












          • $begingroup$
            Check my answer!
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:15
















          $begingroup$
          How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
          $endgroup$
          – Pedro Gomes
          Jan 18 at 19:39




          $begingroup$
          How do I know that if I take a set in $mathscr{C}$, it is in $mathscr{R}$?
          $endgroup$
          – Pedro Gomes
          Jan 18 at 19:39












          $begingroup$
          I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Jan 18 at 21:23






          $begingroup$
          I think $Asetminus B$ is closed in $mathscr{R}$ once $mathscr{R}$ is a ring. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Jan 18 at 21:23














          $begingroup$
          Yes, but is it in $mathscr{C}$?
          $endgroup$
          – greedoid
          Jan 19 at 18:06






          $begingroup$
          Yes, but is it in $mathscr{C}$?
          $endgroup$
          – greedoid
          Jan 19 at 18:06














          $begingroup$
          Check my answer!
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:15




          $begingroup$
          Check my answer!
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:15











          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23


















          0












          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23
















          0












          0








          0





          $begingroup$

          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.






          share|cite|improve this answer









          $endgroup$



          After greedoid answer I am able to post an answer to the question. So we have 4 cases:



          1: $Ainmathscr{R}$ and $Binmathscr{R}$ then $Acup Binmathscr{R}implies Acup Binmathscr{C}$



          2: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c=A^ccap B^c=B^ccap A^c=B^csetminus Ainmathscr{R}{implies} Acup Binmathscr{C}$



          3: $A^cinmathscr{R}$ and $B^cinmathscr{R}$ then $(Acup B)^c= A^ccap B^cinmathscr{R}stackrel{}{implies} Acup Binmathscr{C}$



          It was intended to show that either $Acup Binmathscr{R}$ or $(Acup B)^cinmathscr{R}$,which are the conditions defined in $mathscr{C}$ for $Acup B$ to belong to $mathscr{C}$. The remaining sections of the exercise are about the same reasoning.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 18:14









          Pedro GomesPedro Gomes

          1,8262721




          1,8262721












          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23




















          • $begingroup$
            Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
            $endgroup$
            – greedoid
            Jan 19 at 18:20












          • $begingroup$
            Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
            $endgroup$
            – Pedro Gomes
            Jan 19 at 18:23


















          $begingroup$
          Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
          $endgroup$
          – greedoid
          Jan 19 at 18:20






          $begingroup$
          Is this true: $Ainmathscr{R}$ and $B^cinmathscr{R}$ then $B^csetminus Ainmathscr{R}$
          $endgroup$
          – greedoid
          Jan 19 at 18:20














          $begingroup$
          Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:23






          $begingroup$
          Yes it is. Once rings are closed for finite $setminus$ and we admit that $B^cinmathscr{R}$.It does not mean $Binmathscr{R}$ once the latter is a ring.
          $endgroup$
          – Pedro Gomes
          Jan 19 at 18:23




















          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%2f3078656%2fmathscrc-e-subset-omegae-in-mathscrr-textor-omega-e-in-mathscrr%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

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
          Read more

          Antonio Litta Visconti Arese

          Image
          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
          Read more