Is $∅ ∈ ∅ $ true?

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$begingroup$


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 $emptyset,emptyset,emptyset$ ?










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    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 $emptyset,emptyset,emptyset$ ?










    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 $emptyset,emptyset,emptyset$ ?










      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






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      edited Jan 21 at 9:19









      Eevee Trainer

      5,9471936




      5,9471936










      asked Jan 21 at 3:34









      J.SJ.S

      605




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          3 Answers
          3






          active

          oldest

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          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









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              3 Answers
              3






              active

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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

                  562




                  562





















                      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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