If a set is closed under countable unions, is it closed under countable intersections?

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I am trying to think through the intuition of DeMorgan's laws. If I have a set $Omega$ and a sequence of subsets ($A_n$)=$X_1$, $X_2$, ... how can I know that given the fact $Omega$ is closed under countable unions, $Omega$ is closed under countable intersections, or vice-versa.



Moreover - given the answer to the above, is it enough to show that $Omega$ is closed under countable intersections OR countable unions as part of proving that $Omega$ is a $sigma$-algebra?



Thanks in advance!






measure-theory elementary-set-theory







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    You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
    – Saucy O'Path
    6 hours ago











  • To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
    – Holo
    6 hours ago














up vote
2
down vote

favorite












I am trying to think through the intuition of DeMorgan's laws. If I have a set $Omega$ and a sequence of subsets ($A_n$)=$X_1$, $X_2$, ... how can I know that given the fact $Omega$ is closed under countable unions, $Omega$ is closed under countable intersections, or vice-versa.



Moreover - given the answer to the above, is it enough to show that $Omega$ is closed under countable intersections OR countable unions as part of proving that $Omega$ is a $sigma$-algebra?



Thanks in advance!






measure-theory elementary-set-theory







share|cite|improve this question









New contributor




pestopasta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2




    You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
    – Saucy O'Path
    6 hours ago











  • To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
    – Holo
    6 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am trying to think through the intuition of DeMorgan's laws. If I have a set $Omega$ and a sequence of subsets ($A_n$)=$X_1$, $X_2$, ... how can I know that given the fact $Omega$ is closed under countable unions, $Omega$ is closed under countable intersections, or vice-versa.



Moreover - given the answer to the above, is it enough to show that $Omega$ is closed under countable intersections OR countable unions as part of proving that $Omega$ is a $sigma$-algebra?



Thanks in advance!






measure-theory elementary-set-theory







share|cite|improve this question









New contributor




pestopasta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am trying to think through the intuition of DeMorgan's laws. If I have a set $Omega$ and a sequence of subsets ($A_n$)=$X_1$, $X_2$, ... how can I know that given the fact $Omega$ is closed under countable unions, $Omega$ is closed under countable intersections, or vice-versa.



Moreover - given the answer to the above, is it enough to show that $Omega$ is closed under countable intersections OR countable unions as part of proving that $Omega$ is a $sigma$-algebra?



Thanks in advance!






measure-theory elementary-set-theory




measure-theory elementary-set-theory






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pestopasta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 6 hours ago





















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Check out our Code of Conduct.







  • 2




    You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
    – Saucy O'Path
    6 hours ago











  • To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
    – Holo
    6 hours ago












  • 2




    You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
    – Saucy O'Path
    6 hours ago











  • To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
    – Holo
    6 hours ago







2




2




You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
– Saucy O'Path
6 hours ago





You can't know it, because families of sets can be closed under countable union but not under countable intersections (and viceversa). Most topologies, in fact, are like that.
– Saucy O'Path
6 hours ago













To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
– Holo
6 hours ago




To answer the question in the title, consider all the intervals of form $[0,h)$, where $h>0$, then, because it is linearly ordered under $subseteq$, it is closed under arbitrary union(which is stronger then countable). Now take all the members in the form of $[0,1/n)$ for natural $n$, the intersection is the singleton $0$
– Holo
6 hours ago










4 Answers
4






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up vote
2
down vote



accepted










It is false. Consider as $Omega$ the family of subsets of $mathbbN$ whose complement is finite plus the empty set.



Then $Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $Omega$: take $A_n=mathbbNsetminus1,2,dots,n$; then $bigcap_n A_n=0notinOmega$.



Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).






share|cite|improve this answer



























    up vote
    3
    down vote













    For a small example:



    $mathcalF=1,2,1,2$ is closed under unions... (Each of $1cup1,1cup2,1cup1,2,dots$ etc... are elements of $mathcalF$)



    ...however it is not closed under intersections. ($1cap2=emptysetnotinmathcalF$)






    share|cite|improve this answer



























      up vote
      1
      down vote













      I believe it is if it is also closed under complements.



      Suppose $A in Omega$ and $B in Omega$.



      Then $A^c in Omega$ and $B^c in Omega$.



      Then $(A^c cup B^c)^c = A cap B in Omega$






      share|cite|improve this answer



























        up vote
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        down vote













        Consider the family of sets of natural numbers with at least seventeen elements.



        The union of any number of such sets still has at least seventeen elements. But the intersection ...






        share|cite|improve this answer




















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          It is false. Consider as $Omega$ the family of subsets of $mathbbN$ whose complement is finite plus the empty set.



          Then $Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $Omega$: take $A_n=mathbbNsetminus1,2,dots,n$; then $bigcap_n A_n=0notinOmega$.



          Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).






          share|cite|improve this answer
























            up vote
            2
            down vote



            accepted










            It is false. Consider as $Omega$ the family of subsets of $mathbbN$ whose complement is finite plus the empty set.



            Then $Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $Omega$: take $A_n=mathbbNsetminus1,2,dots,n$; then $bigcap_n A_n=0notinOmega$.



            Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).






            share|cite|improve this answer






















              up vote
              2
              down vote



              accepted







              up vote
              2
              down vote



              accepted






              It is false. Consider as $Omega$ the family of subsets of $mathbbN$ whose complement is finite plus the empty set.



              Then $Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $Omega$: take $A_n=mathbbNsetminus1,2,dots,n$; then $bigcap_n A_n=0notinOmega$.



              Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).






              share|cite|improve this answer












              It is false. Consider as $Omega$ the family of subsets of $mathbbN$ whose complement is finite plus the empty set.



              Then $Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $Omega$: take $A_n=mathbbNsetminus1,2,dots,n$; then $bigcap_n A_n=0notinOmega$.



              Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 6 hours ago









              egreg

              167k1280189




              167k1280189




















                  up vote
                  3
                  down vote













                  For a small example:



                  $mathcalF=1,2,1,2$ is closed under unions... (Each of $1cup1,1cup2,1cup1,2,dots$ etc... are elements of $mathcalF$)



                  ...however it is not closed under intersections. ($1cap2=emptysetnotinmathcalF$)






                  share|cite|improve this answer
























                    up vote
                    3
                    down vote













                    For a small example:



                    $mathcalF=1,2,1,2$ is closed under unions... (Each of $1cup1,1cup2,1cup1,2,dots$ etc... are elements of $mathcalF$)



                    ...however it is not closed under intersections. ($1cap2=emptysetnotinmathcalF$)






                    share|cite|improve this answer






















                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      For a small example:



                      $mathcalF=1,2,1,2$ is closed under unions... (Each of $1cup1,1cup2,1cup1,2,dots$ etc... are elements of $mathcalF$)



                      ...however it is not closed under intersections. ($1cap2=emptysetnotinmathcalF$)






                      share|cite|improve this answer












                      For a small example:



                      $mathcalF=1,2,1,2$ is closed under unions... (Each of $1cup1,1cup2,1cup1,2,dots$ etc... are elements of $mathcalF$)



                      ...however it is not closed under intersections. ($1cap2=emptysetnotinmathcalF$)







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 6 hours ago









                      JMoravitz

                      45k33582




                      45k33582




















                          up vote
                          1
                          down vote













                          I believe it is if it is also closed under complements.



                          Suppose $A in Omega$ and $B in Omega$.



                          Then $A^c in Omega$ and $B^c in Omega$.



                          Then $(A^c cup B^c)^c = A cap B in Omega$






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote













                            I believe it is if it is also closed under complements.



                            Suppose $A in Omega$ and $B in Omega$.



                            Then $A^c in Omega$ and $B^c in Omega$.



                            Then $(A^c cup B^c)^c = A cap B in Omega$






                            share|cite|improve this answer






















                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              I believe it is if it is also closed under complements.



                              Suppose $A in Omega$ and $B in Omega$.



                              Then $A^c in Omega$ and $B^c in Omega$.



                              Then $(A^c cup B^c)^c = A cap B in Omega$






                              share|cite|improve this answer












                              I believe it is if it is also closed under complements.



                              Suppose $A in Omega$ and $B in Omega$.



                              Then $A^c in Omega$ and $B^c in Omega$.



                              Then $(A^c cup B^c)^c = A cap B in Omega$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 6 hours ago









                              Juanma Eloy

                              4841414




                              4841414




















                                  up vote
                                  1
                                  down vote













                                  Consider the family of sets of natural numbers with at least seventeen elements.



                                  The union of any number of such sets still has at least seventeen elements. But the intersection ...






                                  share|cite|improve this answer
























                                    up vote
                                    1
                                    down vote













                                    Consider the family of sets of natural numbers with at least seventeen elements.



                                    The union of any number of such sets still has at least seventeen elements. But the intersection ...






                                    share|cite|improve this answer






















                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      Consider the family of sets of natural numbers with at least seventeen elements.



                                      The union of any number of such sets still has at least seventeen elements. But the intersection ...






                                      share|cite|improve this answer












                                      Consider the family of sets of natural numbers with at least seventeen elements.



                                      The union of any number of such sets still has at least seventeen elements. But the intersection ...







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 6 hours ago









                                      Noah Schweber

                                      113k9143267




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