If $x$ is a set then $cap x$ is a set












1












$begingroup$


Consider the following problem:



(*) If $A$ is a non-empty set, then $cap A$ is a set.



But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?



What did I do wrong?






elementary-set-theory logic







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Consider the following problem:



    (*) If $A$ is a non-empty set, then $cap A$ is a set.



    But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?



    What did I do wrong?






    elementary-set-theory logic







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the following problem:



      (*) If $A$ is a non-empty set, then $cap A$ is a set.



      But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?



      What did I do wrong?






      elementary-set-theory logic







      share|cite|improve this question











      $endgroup$




      Consider the following problem:



      (*) If $A$ is a non-empty set, then $cap A$ is a set.



      But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?



      What did I do wrong?






      elementary-set-theory logic




      elementary-set-theory logic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 12:29









      Andrés E. Caicedo

      65.6k8159250




      65.6k8159250










      asked Dec 6 '18 at 7:15









      bbwbbw

      50039




      50039






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
          definition gives $emptyset$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:24












          • $begingroup$
            If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:38








          • 1




            $begingroup$
            @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:40





















          4












          $begingroup$

          If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank! but if $A$ is non-empty, is there a problem with my proposed definition?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:36










          • $begingroup$
            Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
            $endgroup$
            – Carlos Jiménez
            Dec 6 '18 at 7:43



















          -1












          $begingroup$

          Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
            definition gives $emptyset$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:24












            • $begingroup$
              If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:38








            • 1




              $begingroup$
              @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:40


















            4












            $begingroup$

            Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
            definition gives $emptyset$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:24












            • $begingroup$
              If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:38








            • 1




              $begingroup$
              @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:40
















            4












            4








            4





            $begingroup$

            Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
            definition gives $emptyset$.






            share|cite|improve this answer









            $endgroup$



            Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
            definition gives $emptyset$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 7:18









            Lord Shark the UnknownLord Shark the Unknown

            105k1160133




            105k1160133












            • $begingroup$
              Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:24












            • $begingroup$
              If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:38








            • 1




              $begingroup$
              @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:40




















            • $begingroup$
              Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:24












            • $begingroup$
              If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:38








            • 1




              $begingroup$
              @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
              $endgroup$
              – Arthur
              Dec 6 '18 at 7:40


















            $begingroup$
            Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:24






            $begingroup$
            Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:24














            $begingroup$
            If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:38






            $begingroup$
            If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:38






            1




            1




            $begingroup$
            @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:40






            $begingroup$
            @bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
            $endgroup$
            – Arthur
            Dec 6 '18 at 7:40













            4












            $begingroup$

            If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank! but if $A$ is non-empty, is there a problem with my proposed definition?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:36










            • $begingroup$
              Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
              $endgroup$
              – Carlos Jiménez
              Dec 6 '18 at 7:43
















            4












            $begingroup$

            If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank! but if $A$ is non-empty, is there a problem with my proposed definition?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:36










            • $begingroup$
              Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
              $endgroup$
              – Carlos Jiménez
              Dec 6 '18 at 7:43














            4












            4








            4





            $begingroup$

            If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.






            share|cite|improve this answer









            $endgroup$



            If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 7:24









            Carlos JiménezCarlos Jiménez

            2,4051620




            2,4051620












            • $begingroup$
              thank! but if $A$ is non-empty, is there a problem with my proposed definition?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:36










            • $begingroup$
              Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
              $endgroup$
              – Carlos Jiménez
              Dec 6 '18 at 7:43


















            • $begingroup$
              thank! but if $A$ is non-empty, is there a problem with my proposed definition?
              $endgroup$
              – bbw
              Dec 6 '18 at 7:36










            • $begingroup$
              Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
              $endgroup$
              – Carlos Jiménez
              Dec 6 '18 at 7:43
















            $begingroup$
            thank! but if $A$ is non-empty, is there a problem with my proposed definition?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:36




            $begingroup$
            thank! but if $A$ is non-empty, is there a problem with my proposed definition?
            $endgroup$
            – bbw
            Dec 6 '18 at 7:36












            $begingroup$
            Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
            $endgroup$
            – Carlos Jiménez
            Dec 6 '18 at 7:43




            $begingroup$
            Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
            $endgroup$
            – Carlos Jiménez
            Dec 6 '18 at 7:43











            -1












            $begingroup$

            Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.






            share|cite|improve this answer









            $endgroup$


















              -1












              $begingroup$

              Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.






              share|cite|improve this answer









              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.






                share|cite|improve this answer









                $endgroup$



                Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 10:32









                DanielWainfleetDanielWainfleet

                35.3k31648




                35.3k31648






























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