Working with the case $Omega$ is countable and $Omega$ is uncountable for a singleton set generator...












3














Let $Omega neq varnothing$. Let $mathcal{E}:={ {omega} : omega in Omega}$



Show:



i) If $Omega$ is countable, then $sigma(mathcal{E})=2^{Omega}$.



ii) If $Omega$ is uncountable, then: if $A in sigma(mathcal{E}) iff A$ or $A^{C}$ is countable



My ideas:



i) I think (i) is rather trivial.



For $sigma(mathcal{E})subseteq 2^{Omega}$. Let $A in sigma(mathcal{E})$. Then $A$ can be written as a countable union of singletons which are by definition $subseteq Omega$ and therefore $in 2^{Omega}$.



$2^{Omega} subseteq sigma(mathcal{E})$ is trivial.



ii) I do not really know where to begin with an uncountable $Omega$. Any tips?






real-analysis probability measure-theory







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    3














    Let $Omega neq varnothing$. Let $mathcal{E}:={ {omega} : omega in Omega}$



    Show:



    i) If $Omega$ is countable, then $sigma(mathcal{E})=2^{Omega}$.



    ii) If $Omega$ is uncountable, then: if $A in sigma(mathcal{E}) iff A$ or $A^{C}$ is countable



    My ideas:



    i) I think (i) is rather trivial.



    For $sigma(mathcal{E})subseteq 2^{Omega}$. Let $A in sigma(mathcal{E})$. Then $A$ can be written as a countable union of singletons which are by definition $subseteq Omega$ and therefore $in 2^{Omega}$.



    $2^{Omega} subseteq sigma(mathcal{E})$ is trivial.



    ii) I do not really know where to begin with an uncountable $Omega$. Any tips?






    real-analysis probability measure-theory







    share|cite|improve this question

























      3












      3








      3







      Let $Omega neq varnothing$. Let $mathcal{E}:={ {omega} : omega in Omega}$



      Show:



      i) If $Omega$ is countable, then $sigma(mathcal{E})=2^{Omega}$.



      ii) If $Omega$ is uncountable, then: if $A in sigma(mathcal{E}) iff A$ or $A^{C}$ is countable



      My ideas:



      i) I think (i) is rather trivial.



      For $sigma(mathcal{E})subseteq 2^{Omega}$. Let $A in sigma(mathcal{E})$. Then $A$ can be written as a countable union of singletons which are by definition $subseteq Omega$ and therefore $in 2^{Omega}$.



      $2^{Omega} subseteq sigma(mathcal{E})$ is trivial.



      ii) I do not really know where to begin with an uncountable $Omega$. Any tips?






      real-analysis probability measure-theory







      share|cite|improve this question













      Let $Omega neq varnothing$. Let $mathcal{E}:={ {omega} : omega in Omega}$



      Show:



      i) If $Omega$ is countable, then $sigma(mathcal{E})=2^{Omega}$.



      ii) If $Omega$ is uncountable, then: if $A in sigma(mathcal{E}) iff A$ or $A^{C}$ is countable



      My ideas:



      i) I think (i) is rather trivial.



      For $sigma(mathcal{E})subseteq 2^{Omega}$. Let $A in sigma(mathcal{E})$. Then $A$ can be written as a countable union of singletons which are by definition $subseteq Omega$ and therefore $in 2^{Omega}$.



      $2^{Omega} subseteq sigma(mathcal{E})$ is trivial.



      ii) I do not really know where to begin with an uncountable $Omega$. Any tips?






      real-analysis probability measure-theory




      real-analysis probability measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 '18 at 22:24









      SABOY

      549311




      549311






















          1 Answer
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          Here's a suggestion. By definition $sigma(mathcal{E})$ is the smallest sigma algebra containing $mathcal{E}$. Thus if we want to show that $newcommandcalA{mathcal{A}}newcommandcalB{mathcal{B}}newcommandcalE{mathcal{E}}sigma(calE)=calA$, a common strategy is to show that $calA$ is a $sigma$-algebra and that $calEsubseteq calA$ and if $calB$ is any $sigma$-algebra containing $calE$ then $calAsubseteq calB$.



          In your case, $calA$ will be the set of all subsets $A$ of $Omega$ such that either $A$ is countable or its complement is countable. Then to show $sigma(calE)=calA$ you need to show:





          1. $calA$ is a $sigma$-algebra,


          2. $calEsubseteq calA$, and

          3. if $calB$ is a $sigma$-algebra containing $calE$ then $calAsubseteq calB$.


          Can you show these things?



          Side note



          In case its unclear why these properties show that $sigma(calE)=calA$, note that the third property immediately gives us $calAsubseteq sigma(calE)$ (since $sigma(calE)$ is a $sigma$-algebra containing $calE$), but we also have by definition that $sigma(calE)$ is the intersection of all $sigma$-algebras containing $calE$, which includes $calA$ by properties 1 and 2, so $sigma(calE)subseteq calA$ as well.



          Edit



          Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.



          Hint: How can a countable set be written in terms of elements of $calE$?




          Let $calB$ be a sigma algebra containing $calE$. First I claim any countable subset of $Omega$ is in $calB$, since if $AsubsetOmega$ is countable, then $A=bigcup_{xin A} {x}$, which is a countable union of elements of $calE$, and therefore in $calB$. If $A$ has a countable complement on the other hand, then its complement is in $calB$ by the argument above, and so it is in $calB$ as well. Thus $calB$ contains $calA$.







          share|cite|improve this answer























          • Thanks! I am stuck on step 3 though. How can I prove this?
            – SABOY
            Nov 25 '18 at 23:00











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          3














          Here's a suggestion. By definition $sigma(mathcal{E})$ is the smallest sigma algebra containing $mathcal{E}$. Thus if we want to show that $newcommandcalA{mathcal{A}}newcommandcalB{mathcal{B}}newcommandcalE{mathcal{E}}sigma(calE)=calA$, a common strategy is to show that $calA$ is a $sigma$-algebra and that $calEsubseteq calA$ and if $calB$ is any $sigma$-algebra containing $calE$ then $calAsubseteq calB$.



          In your case, $calA$ will be the set of all subsets $A$ of $Omega$ such that either $A$ is countable or its complement is countable. Then to show $sigma(calE)=calA$ you need to show:





          1. $calA$ is a $sigma$-algebra,


          2. $calEsubseteq calA$, and

          3. if $calB$ is a $sigma$-algebra containing $calE$ then $calAsubseteq calB$.


          Can you show these things?



          Side note



          In case its unclear why these properties show that $sigma(calE)=calA$, note that the third property immediately gives us $calAsubseteq sigma(calE)$ (since $sigma(calE)$ is a $sigma$-algebra containing $calE$), but we also have by definition that $sigma(calE)$ is the intersection of all $sigma$-algebras containing $calE$, which includes $calA$ by properties 1 and 2, so $sigma(calE)subseteq calA$ as well.



          Edit



          Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.



          Hint: How can a countable set be written in terms of elements of $calE$?




          Let $calB$ be a sigma algebra containing $calE$. First I claim any countable subset of $Omega$ is in $calB$, since if $AsubsetOmega$ is countable, then $A=bigcup_{xin A} {x}$, which is a countable union of elements of $calE$, and therefore in $calB$. If $A$ has a countable complement on the other hand, then its complement is in $calB$ by the argument above, and so it is in $calB$ as well. Thus $calB$ contains $calA$.







          share|cite|improve this answer























          • Thanks! I am stuck on step 3 though. How can I prove this?
            – SABOY
            Nov 25 '18 at 23:00
















          3














          Here's a suggestion. By definition $sigma(mathcal{E})$ is the smallest sigma algebra containing $mathcal{E}$. Thus if we want to show that $newcommandcalA{mathcal{A}}newcommandcalB{mathcal{B}}newcommandcalE{mathcal{E}}sigma(calE)=calA$, a common strategy is to show that $calA$ is a $sigma$-algebra and that $calEsubseteq calA$ and if $calB$ is any $sigma$-algebra containing $calE$ then $calAsubseteq calB$.



          In your case, $calA$ will be the set of all subsets $A$ of $Omega$ such that either $A$ is countable or its complement is countable. Then to show $sigma(calE)=calA$ you need to show:





          1. $calA$ is a $sigma$-algebra,


          2. $calEsubseteq calA$, and

          3. if $calB$ is a $sigma$-algebra containing $calE$ then $calAsubseteq calB$.


          Can you show these things?



          Side note



          In case its unclear why these properties show that $sigma(calE)=calA$, note that the third property immediately gives us $calAsubseteq sigma(calE)$ (since $sigma(calE)$ is a $sigma$-algebra containing $calE$), but we also have by definition that $sigma(calE)$ is the intersection of all $sigma$-algebras containing $calE$, which includes $calA$ by properties 1 and 2, so $sigma(calE)subseteq calA$ as well.



          Edit



          Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.



          Hint: How can a countable set be written in terms of elements of $calE$?




          Let $calB$ be a sigma algebra containing $calE$. First I claim any countable subset of $Omega$ is in $calB$, since if $AsubsetOmega$ is countable, then $A=bigcup_{xin A} {x}$, which is a countable union of elements of $calE$, and therefore in $calB$. If $A$ has a countable complement on the other hand, then its complement is in $calB$ by the argument above, and so it is in $calB$ as well. Thus $calB$ contains $calA$.







          share|cite|improve this answer























          • Thanks! I am stuck on step 3 though. How can I prove this?
            – SABOY
            Nov 25 '18 at 23:00














          3












          3








          3






          Here's a suggestion. By definition $sigma(mathcal{E})$ is the smallest sigma algebra containing $mathcal{E}$. Thus if we want to show that $newcommandcalA{mathcal{A}}newcommandcalB{mathcal{B}}newcommandcalE{mathcal{E}}sigma(calE)=calA$, a common strategy is to show that $calA$ is a $sigma$-algebra and that $calEsubseteq calA$ and if $calB$ is any $sigma$-algebra containing $calE$ then $calAsubseteq calB$.



          In your case, $calA$ will be the set of all subsets $A$ of $Omega$ such that either $A$ is countable or its complement is countable. Then to show $sigma(calE)=calA$ you need to show:





          1. $calA$ is a $sigma$-algebra,


          2. $calEsubseteq calA$, and

          3. if $calB$ is a $sigma$-algebra containing $calE$ then $calAsubseteq calB$.


          Can you show these things?



          Side note



          In case its unclear why these properties show that $sigma(calE)=calA$, note that the third property immediately gives us $calAsubseteq sigma(calE)$ (since $sigma(calE)$ is a $sigma$-algebra containing $calE$), but we also have by definition that $sigma(calE)$ is the intersection of all $sigma$-algebras containing $calE$, which includes $calA$ by properties 1 and 2, so $sigma(calE)subseteq calA$ as well.



          Edit



          Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.



          Hint: How can a countable set be written in terms of elements of $calE$?




          Let $calB$ be a sigma algebra containing $calE$. First I claim any countable subset of $Omega$ is in $calB$, since if $AsubsetOmega$ is countable, then $A=bigcup_{xin A} {x}$, which is a countable union of elements of $calE$, and therefore in $calB$. If $A$ has a countable complement on the other hand, then its complement is in $calB$ by the argument above, and so it is in $calB$ as well. Thus $calB$ contains $calA$.







          share|cite|improve this answer














          Here's a suggestion. By definition $sigma(mathcal{E})$ is the smallest sigma algebra containing $mathcal{E}$. Thus if we want to show that $newcommandcalA{mathcal{A}}newcommandcalB{mathcal{B}}newcommandcalE{mathcal{E}}sigma(calE)=calA$, a common strategy is to show that $calA$ is a $sigma$-algebra and that $calEsubseteq calA$ and if $calB$ is any $sigma$-algebra containing $calE$ then $calAsubseteq calB$.



          In your case, $calA$ will be the set of all subsets $A$ of $Omega$ such that either $A$ is countable or its complement is countable. Then to show $sigma(calE)=calA$ you need to show:





          1. $calA$ is a $sigma$-algebra,


          2. $calEsubseteq calA$, and

          3. if $calB$ is a $sigma$-algebra containing $calE$ then $calAsubseteq calB$.


          Can you show these things?



          Side note



          In case its unclear why these properties show that $sigma(calE)=calA$, note that the third property immediately gives us $calAsubseteq sigma(calE)$ (since $sigma(calE)$ is a $sigma$-algebra containing $calE$), but we also have by definition that $sigma(calE)$ is the intersection of all $sigma$-algebras containing $calE$, which includes $calA$ by properties 1 and 2, so $sigma(calE)subseteq calA$ as well.



          Edit



          Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.



          Hint: How can a countable set be written in terms of elements of $calE$?




          Let $calB$ be a sigma algebra containing $calE$. First I claim any countable subset of $Omega$ is in $calB$, since if $AsubsetOmega$ is countable, then $A=bigcup_{xin A} {x}$, which is a countable union of elements of $calE$, and therefore in $calB$. If $A$ has a countable complement on the other hand, then its complement is in $calB$ by the argument above, and so it is in $calB$ as well. Thus $calB$ contains $calA$.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 '18 at 23:05

























          answered Nov 25 '18 at 22:38









          jgon

          13k21941




          13k21941












          • Thanks! I am stuck on step 3 though. How can I prove this?
            – SABOY
            Nov 25 '18 at 23:00


















          • Thanks! I am stuck on step 3 though. How can I prove this?
            – SABOY
            Nov 25 '18 at 23:00
















          Thanks! I am stuck on step 3 though. How can I prove this?
          – SABOY
          Nov 25 '18 at 23:00




          Thanks! I am stuck on step 3 though. How can I prove this?
          – SABOY
          Nov 25 '18 at 23:00


















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