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












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 notation







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$endgroup$












  • $begingroup$
    A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
    $endgroup$
    – Trebor
    yesterday
















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 notation







share|cite|improve this question











$endgroup$












  • $begingroup$
    A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
    $endgroup$
    – Trebor
    yesterday














7












7








7


1



$begingroup$


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 notation







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 notation




measure-theory elementary-set-theory notation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 14 at 1:59









YuiTo Cheng

2,54841037




2,54841037










asked Apr 13 at 17:00









Ica SanduIca Sandu

1399




1399












  • $begingroup$
    A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
    $endgroup$
    – Trebor
    yesterday


















  • $begingroup$
    A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
    $endgroup$
    – Trebor
    yesterday
















$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday




$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday










4 Answers
4






active

oldest

votes


















24












$begingroup$

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











$endgroup$





















    6












    $begingroup$

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






    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









      $endgroup$





















        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









        $endgroup$














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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          24












          $begingroup$

          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











          $endgroup$


















            24












            $begingroup$

            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











            $endgroup$
















              24












              24








              24





              $begingroup$

              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 Apr 13 at 18:36

























              answered Apr 13 at 17:03









              CornmanCornman

              3,81021233




              3,81021233























                  6












                  $begingroup$

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






                  share|cite|improve this answer











                  $endgroup$


















                    6












                    $begingroup$

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






                    share|cite|improve this answer











                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, 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 $cup,emptyset $, 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 Apr 15 at 10:39

























                      answered Apr 13 at 18:57









                      CiaPanCiaPan

                      10.4k11248




                      10.4k11248























                          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 Apr 13 at 17:03









                              José Carlos SantosJosé Carlos Santos

                              176k24135244




                              176k24135244























                                  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 Apr 13 at 17:05









                                      MelodyMelody

                                      1,31212




                                      1,31212






























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