Is $∅ ∈ { {∅} }$ true?












9












$begingroup$


If $ {emptyset} ∈ {emptyset,{emptyset}} $ is true, does it mean this $ emptyset in {{emptyset}} $ true ? If it is not, why it is false?



Also, does $ {{emptyset}}$ mean ${emptyset,{emptyset,{emptyset}}}$ ?






elementary-set-theory







share|cite|improve this question











$endgroup$

















    9












    $begingroup$


    If $ {emptyset} ∈ {emptyset,{emptyset}} $ is true, does it mean this $ emptyset in {{emptyset}} $ true ? If it is not, why it is false?



    Also, does $ {{emptyset}}$ mean ${emptyset,{emptyset,{emptyset}}}$ ?






    elementary-set-theory







    share|cite|improve this question











    $endgroup$















      9












      9








      9


      1



      $begingroup$


      If $ {emptyset} ∈ {emptyset,{emptyset}} $ is true, does it mean this $ emptyset in {{emptyset}} $ true ? If it is not, why it is false?



      Also, does $ {{emptyset}}$ mean ${emptyset,{emptyset,{emptyset}}}$ ?






      elementary-set-theory







      share|cite|improve this question











      $endgroup$




      If $ {emptyset} ∈ {emptyset,{emptyset}} $ is true, does it mean this $ emptyset in {{emptyset}} $ true ? If it is not, why it is false?



      Also, does $ {{emptyset}}$ mean ${emptyset,{emptyset,{emptyset}}}$ ?






      elementary-set-theory




      elementary-set-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 9:19









      Eevee Trainer

      5,8011936




      5,8011936










      asked Jan 21 at 3:34









      J.SJ.S

      555




      555






















          3 Answers
          3






          active

          oldest

          votes


















          30












          $begingroup$

          The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$



          The notation ${emptyset, {emptyset}}$ describes a set with exactly two elements.



          The first element is $emptyset.$ The second element is ${emptyset}.$
          Is one of those two elements exactly equal to ${emptyset}$?



          The notation ${ {emptyset}}$ describes a set with one element.
          That element is ${emptyset}.$



          Which element of ${ {emptyset}}$ do you think is exactly equal to $emptyset$?
          Hint: there's only one element you have to check.



          The notation ${emptyset, {emptyset, {emptyset}}}$ again describes a set with two elements.
          One element is $emptyset$ and the other is
          ${emptyset, {emptyset}}.$
          So this is definitely not the same thing as any set that has only one element.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is not an answer but a counter-question.
            $endgroup$
            – rexkogitans
            Jan 21 at 9:36






          • 2




            $begingroup$
            @rexkogitans Also known as a "hint"
            $endgroup$
            – TreFox
            Jan 21 at 19:36



















          2












          $begingroup$

          The set ${{emptyset}}$ is a set with one element: ${emptyset}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).



          Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.



          This also explains why ${{emptyset}} notequiv {emptyset,{emptyset,{emptyset}}}$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Let us begin with naming the sets in the question: The set $B={a}$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B={emptyset}$, which is of course also a set with a single elementy, namely $emptyset$.



            If we put $A$ into a set $C={B}$, then we have $C={{emptyset}}$. It can be easily seen that $a notin C$. Hence, $emptyset notin {{emptyset}}$.



            The set $A={emptyset, b}$ has two elements. Now, let $b={emptyset}$. Then we have the set $A={emptyset, {emptyset}}$. If we now take a look at $b$, then, since $bin A$, also ${emptyset} in A$, which is ${emptyset} in {emptyset, {emptyset}}$.



            To answer if ${{emptyset}}$ means ${emptyset,{emptyset,{emptyset}}}$:
            Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              30












              $begingroup$

              The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$



              The notation ${emptyset, {emptyset}}$ describes a set with exactly two elements.



              The first element is $emptyset.$ The second element is ${emptyset}.$
              Is one of those two elements exactly equal to ${emptyset}$?



              The notation ${ {emptyset}}$ describes a set with one element.
              That element is ${emptyset}.$



              Which element of ${ {emptyset}}$ do you think is exactly equal to $emptyset$?
              Hint: there's only one element you have to check.



              The notation ${emptyset, {emptyset, {emptyset}}}$ again describes a set with two elements.
              One element is $emptyset$ and the other is
              ${emptyset, {emptyset}}.$
              So this is definitely not the same thing as any set that has only one element.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is not an answer but a counter-question.
                $endgroup$
                – rexkogitans
                Jan 21 at 9:36






              • 2




                $begingroup$
                @rexkogitans Also known as a "hint"
                $endgroup$
                – TreFox
                Jan 21 at 19:36
















              30












              $begingroup$

              The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$



              The notation ${emptyset, {emptyset}}$ describes a set with exactly two elements.



              The first element is $emptyset.$ The second element is ${emptyset}.$
              Is one of those two elements exactly equal to ${emptyset}$?



              The notation ${ {emptyset}}$ describes a set with one element.
              That element is ${emptyset}.$



              Which element of ${ {emptyset}}$ do you think is exactly equal to $emptyset$?
              Hint: there's only one element you have to check.



              The notation ${emptyset, {emptyset, {emptyset}}}$ again describes a set with two elements.
              One element is $emptyset$ and the other is
              ${emptyset, {emptyset}}.$
              So this is definitely not the same thing as any set that has only one element.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is not an answer but a counter-question.
                $endgroup$
                – rexkogitans
                Jan 21 at 9:36






              • 2




                $begingroup$
                @rexkogitans Also known as a "hint"
                $endgroup$
                – TreFox
                Jan 21 at 19:36














              30












              30








              30





              $begingroup$

              The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$



              The notation ${emptyset, {emptyset}}$ describes a set with exactly two elements.



              The first element is $emptyset.$ The second element is ${emptyset}.$
              Is one of those two elements exactly equal to ${emptyset}$?



              The notation ${ {emptyset}}$ describes a set with one element.
              That element is ${emptyset}.$



              Which element of ${ {emptyset}}$ do you think is exactly equal to $emptyset$?
              Hint: there's only one element you have to check.



              The notation ${emptyset, {emptyset, {emptyset}}}$ again describes a set with two elements.
              One element is $emptyset$ and the other is
              ${emptyset, {emptyset}}.$
              So this is definitely not the same thing as any set that has only one element.






              share|cite|improve this answer









              $endgroup$



              The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$



              The notation ${emptyset, {emptyset}}$ describes a set with exactly two elements.



              The first element is $emptyset.$ The second element is ${emptyset}.$
              Is one of those two elements exactly equal to ${emptyset}$?



              The notation ${ {emptyset}}$ describes a set with one element.
              That element is ${emptyset}.$



              Which element of ${ {emptyset}}$ do you think is exactly equal to $emptyset$?
              Hint: there's only one element you have to check.



              The notation ${emptyset, {emptyset, {emptyset}}}$ again describes a set with two elements.
              One element is $emptyset$ and the other is
              ${emptyset, {emptyset}}.$
              So this is definitely not the same thing as any set that has only one element.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 21 at 3:51









              David KDavid K

              53.9k342116




              53.9k342116












              • $begingroup$
                This is not an answer but a counter-question.
                $endgroup$
                – rexkogitans
                Jan 21 at 9:36






              • 2




                $begingroup$
                @rexkogitans Also known as a "hint"
                $endgroup$
                – TreFox
                Jan 21 at 19:36


















              • $begingroup$
                This is not an answer but a counter-question.
                $endgroup$
                – rexkogitans
                Jan 21 at 9:36






              • 2




                $begingroup$
                @rexkogitans Also known as a "hint"
                $endgroup$
                – TreFox
                Jan 21 at 19:36
















              $begingroup$
              This is not an answer but a counter-question.
              $endgroup$
              – rexkogitans
              Jan 21 at 9:36




              $begingroup$
              This is not an answer but a counter-question.
              $endgroup$
              – rexkogitans
              Jan 21 at 9:36




              2




              2




              $begingroup$
              @rexkogitans Also known as a "hint"
              $endgroup$
              – TreFox
              Jan 21 at 19:36




              $begingroup$
              @rexkogitans Also known as a "hint"
              $endgroup$
              – TreFox
              Jan 21 at 19:36











              2












              $begingroup$

              The set ${{emptyset}}$ is a set with one element: ${emptyset}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).



              Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.



              This also explains why ${{emptyset}} notequiv {emptyset,{emptyset,{emptyset}}}$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The set ${{emptyset}}$ is a set with one element: ${emptyset}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).



                Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.



                This also explains why ${{emptyset}} notequiv {emptyset,{emptyset,{emptyset}}}$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The set ${{emptyset}}$ is a set with one element: ${emptyset}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).



                  Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.



                  This also explains why ${{emptyset}} notequiv {emptyset,{emptyset,{emptyset}}}$.






                  share|cite|improve this answer









                  $endgroup$



                  The set ${{emptyset}}$ is a set with one element: ${emptyset}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).



                  Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.



                  This also explains why ${{emptyset}} notequiv {emptyset,{emptyset,{emptyset}}}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 9:08









                  AdrianAdrian

                  462




                  462























                      0












                      $begingroup$

                      Let us begin with naming the sets in the question: The set $B={a}$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B={emptyset}$, which is of course also a set with a single elementy, namely $emptyset$.



                      If we put $A$ into a set $C={B}$, then we have $C={{emptyset}}$. It can be easily seen that $a notin C$. Hence, $emptyset notin {{emptyset}}$.



                      The set $A={emptyset, b}$ has two elements. Now, let $b={emptyset}$. Then we have the set $A={emptyset, {emptyset}}$. If we now take a look at $b$, then, since $bin A$, also ${emptyset} in A$, which is ${emptyset} in {emptyset, {emptyset}}$.



                      To answer if ${{emptyset}}$ means ${emptyset,{emptyset,{emptyset}}}$:
                      Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Let us begin with naming the sets in the question: The set $B={a}$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B={emptyset}$, which is of course also a set with a single elementy, namely $emptyset$.



                        If we put $A$ into a set $C={B}$, then we have $C={{emptyset}}$. It can be easily seen that $a notin C$. Hence, $emptyset notin {{emptyset}}$.



                        The set $A={emptyset, b}$ has two elements. Now, let $b={emptyset}$. Then we have the set $A={emptyset, {emptyset}}$. If we now take a look at $b$, then, since $bin A$, also ${emptyset} in A$, which is ${emptyset} in {emptyset, {emptyset}}$.



                        To answer if ${{emptyset}}$ means ${emptyset,{emptyset,{emptyset}}}$:
                        Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Let us begin with naming the sets in the question: The set $B={a}$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B={emptyset}$, which is of course also a set with a single elementy, namely $emptyset$.



                          If we put $A$ into a set $C={B}$, then we have $C={{emptyset}}$. It can be easily seen that $a notin C$. Hence, $emptyset notin {{emptyset}}$.



                          The set $A={emptyset, b}$ has two elements. Now, let $b={emptyset}$. Then we have the set $A={emptyset, {emptyset}}$. If we now take a look at $b$, then, since $bin A$, also ${emptyset} in A$, which is ${emptyset} in {emptyset, {emptyset}}$.



                          To answer if ${{emptyset}}$ means ${emptyset,{emptyset,{emptyset}}}$:
                          Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).






                          share|cite|improve this answer









                          $endgroup$



                          Let us begin with naming the sets in the question: The set $B={a}$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B={emptyset}$, which is of course also a set with a single elementy, namely $emptyset$.



                          If we put $A$ into a set $C={B}$, then we have $C={{emptyset}}$. It can be easily seen that $a notin C$. Hence, $emptyset notin {{emptyset}}$.



                          The set $A={emptyset, b}$ has two elements. Now, let $b={emptyset}$. Then we have the set $A={emptyset, {emptyset}}$. If we now take a look at $b$, then, since $bin A$, also ${emptyset} in A$, which is ${emptyset} in {emptyset, {emptyset}}$.



                          To answer if ${{emptyset}}$ means ${emptyset,{emptyset,{emptyset}}}$:
                          Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 21 at 9:36









                          rexkogitansrexkogitans

                          1515




                          1515






























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