When is a collection of sets closed under union?












0












$begingroup$


I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".



A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:




  • $Omega in F$


  • $F$ is closed under complement


  • $F$ is closed under union


Now, consider the following example:



Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?



I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.






probability elementary-set-theory







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  • 2




    $begingroup$
    $sigma$-algebras are only required to be closed under countable unions.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 8:42










  • $begingroup$
    slight change to title; a collection of sets can be closed under union, not the set itself;
    $endgroup$
    – Chris
    Jan 6 at 9:09










  • $begingroup$
    You're right! Thanks, I changed it.
    $endgroup$
    – André
    Jan 6 at 9:10
















0












$begingroup$


I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".



A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:




  • $Omega in F$


  • $F$ is closed under complement


  • $F$ is closed under union


Now, consider the following example:



Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?



I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.






probability elementary-set-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $sigma$-algebras are only required to be closed under countable unions.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 8:42










  • $begingroup$
    slight change to title; a collection of sets can be closed under union, not the set itself;
    $endgroup$
    – Chris
    Jan 6 at 9:09










  • $begingroup$
    You're right! Thanks, I changed it.
    $endgroup$
    – André
    Jan 6 at 9:10














0












0








0





$begingroup$


I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".



A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:




  • $Omega in F$


  • $F$ is closed under complement


  • $F$ is closed under union


Now, consider the following example:



Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?



I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.






probability elementary-set-theory







share|cite|improve this question











$endgroup$




I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".



A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:




  • $Omega in F$


  • $F$ is closed under complement


  • $F$ is closed under union


Now, consider the following example:



Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?



I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.






probability elementary-set-theory




probability 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 6 at 9:27









Asaf Karagila♦

306k33437768




306k33437768










asked Jan 6 at 8:30









AndréAndré

63




63








  • 2




    $begingroup$
    $sigma$-algebras are only required to be closed under countable unions.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 8:42










  • $begingroup$
    slight change to title; a collection of sets can be closed under union, not the set itself;
    $endgroup$
    – Chris
    Jan 6 at 9:09










  • $begingroup$
    You're right! Thanks, I changed it.
    $endgroup$
    – André
    Jan 6 at 9:10














  • 2




    $begingroup$
    $sigma$-algebras are only required to be closed under countable unions.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 8:42










  • $begingroup$
    slight change to title; a collection of sets can be closed under union, not the set itself;
    $endgroup$
    – Chris
    Jan 6 at 9:09










  • $begingroup$
    You're right! Thanks, I changed it.
    $endgroup$
    – André
    Jan 6 at 9:10








2




2




$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42




$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42












$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09




$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09












$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10




$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10










2 Answers
2






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$begingroup$

 "$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.



So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    “Closed under union” means that the union of any set of members of $F$ is also a member of $F$.



    In your example $F$ is not closed under union (although it is closed under complement).






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      2












      $begingroup$

       "$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.



      So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

         "$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.



        So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

           "$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.



          So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.






          share|cite|improve this answer











          $endgroup$



           "$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.



          So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 9:20









          Henno Brandsma

          112k348120




          112k348120










          answered Jan 6 at 8:37









          MindlackMindlack

          4,885211




          4,885211























              1












              $begingroup$

              “Closed under union” means that the union of any set of members of $F$ is also a member of $F$.



              In your example $F$ is not closed under union (although it is closed under complement).






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                “Closed under union” means that the union of any set of members of $F$ is also a member of $F$.



                In your example $F$ is not closed under union (although it is closed under complement).






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  “Closed under union” means that the union of any set of members of $F$ is also a member of $F$.



                  In your example $F$ is not closed under union (although it is closed under complement).






                  share|cite|improve this answer









                  $endgroup$



                  “Closed under union” means that the union of any set of members of $F$ is also a member of $F$.



                  In your example $F$ is not closed under union (although it is closed under complement).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 8:39









                  gandalf61gandalf61

                  8,936725




                  8,936725






























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