If the empty set is a subset of every set, why write … ∪ {∅}?












7












$begingroup$


I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $



I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?






measure-theory elementary-set-theory







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $



    I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?






    measure-theory elementary-set-theory







    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $



      I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?






      measure-theory elementary-set-theory







      share|cite|improve this question











      $endgroup$




      I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $



      I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?






      measure-theory elementary-set-theory




      measure-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 45 mins ago









      LarsH

      555624




      555624










      asked 7 hours ago









      Ica SanduIca Sandu

      1379




      1379






















          4 Answers
          4






          active

          oldest

          votes


















          22












          $begingroup$

          It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
          It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.



          Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$






          share|cite|improve this answer











          $endgroup$





















            5












            $begingroup$

            Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).






            share|cite|improve this answer









            $endgroup$





















              5












              $begingroup$

              The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.






              share|cite|improve this answer











              $endgroup$





















                3












                $begingroup$

                It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.



                As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$






                share|cite|improve this answer









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                  4 Answers
                  4






                  active

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                  4 Answers
                  4






                  active

                  oldest

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                  active

                  oldest

                  votes






                  active

                  oldest

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                  22












                  $begingroup$

                  It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
                  It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.



                  Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$






                  share|cite|improve this answer











                  $endgroup$


















                    22












                    $begingroup$

                    It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
                    It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.



                    Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$






                    share|cite|improve this answer











                    $endgroup$
















                      22












                      22








                      22





                      $begingroup$

                      It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
                      It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.



                      Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$






                      share|cite|improve this answer











                      $endgroup$



                      It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
                      It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.



                      Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 6 hours ago

























                      answered 7 hours ago









                      CornmanCornman

                      3,69321231




                      3,69321231























                          5












                          $begingroup$

                          Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).






                          share|cite|improve this answer









                          $endgroup$


















                            5












                            $begingroup$

                            Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).






                            share|cite|improve this answer









                            $endgroup$
















                              5












                              5








                              5





                              $begingroup$

                              Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).






                              share|cite|improve this answer









                              $endgroup$



                              Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 7 hours ago









                              José Carlos SantosJosé Carlos Santos

                              174k23134243




                              174k23134243























                                  5












                                  $begingroup$

                                  The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    5












                                    $begingroup$

                                    The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      5












                                      5








                                      5





                                      $begingroup$

                                      The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.






                                      share|cite|improve this answer











                                      $endgroup$



                                      The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 5 hours ago

























                                      answered 5 hours ago









                                      CiaPanCiaPan

                                      10.3k11248




                                      10.3k11248























                                          3












                                          $begingroup$

                                          It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.



                                          As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$






                                          share|cite|improve this answer









                                          $endgroup$


















                                            3












                                            $begingroup$

                                            It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.



                                            As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$






                                            share|cite|improve this answer









                                            $endgroup$
















                                              3












                                              3








                                              3





                                              $begingroup$

                                              It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.



                                              As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$






                                              share|cite|improve this answer









                                              $endgroup$



                                              It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.



                                              As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 7 hours ago









                                              MelodyMelody

                                              1,21312




                                              1,21312






























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