A family of sets that is not a $sigma$-algebra











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Let $X$ be a not empty set.



I must prove that the family $mathcal{A}$ defined as follows
$$mathcal{A}=bigg{Ainmathcal{P}(X);|;A;text{is finite or $A^c$ is finite}bigg}$$ it is not a $sigma$-algebra if $X$ is infinite.



Obviously $mathcal{A}$ is an algebra. To show that it is not a $sigma$-algebra I reasoned as follows: fixed a sequence ${x_n}_{ninmathbb{N}}$ of elements of $X$ such that $x_ine x_j$ for $ine j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n={x_n}$ for all $ninmathbb{N}$. At this point I consider the sets $A_2,A_4dots ,A_{2k},dots$ where $kinmathbb{N}$; I observe that $A_{2k}inmathcal{A}$ for all $kinmathbb{N}$, but the union $$bigcup_{k=1}^{+infty}A_{2k}=bigg{x_2,x_4,dots,x_{2k},dotsbigg}$$
is infinite. On the other hand
begin{equation}
begin{split}
bigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^c=Xsetminusbigcup_{k=1}^{+infty}A_{2k}=& Xcapbigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^{c}=Xcapbigg[bigcup_{k=1}^{+infty}A_{2k+1}bigg]\
=&bigcup_{k=1}^{+infty}A_{2k+1}cuptilde{A},
end{split}
end{equation}

where $tilde{A}$ it is what remains if $X$ is not countable, if $X$ is countable $tilde{A}=emptyset$. Therefore $cup_k A_{2k}$ and $[cup_k A_{2k}]^c$ are both infinite, then $$bigcup_{n=1}^{+infty} A_nnotinmathcal{A}.$$



it's correct? Thanks!










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  • Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
    – Sorin Tirc
    Nov 18 at 13:52










  • Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
    – Ethan Bolker
    Nov 18 at 13:52















up vote
1
down vote

favorite












Let $X$ be a not empty set.



I must prove that the family $mathcal{A}$ defined as follows
$$mathcal{A}=bigg{Ainmathcal{P}(X);|;A;text{is finite or $A^c$ is finite}bigg}$$ it is not a $sigma$-algebra if $X$ is infinite.



Obviously $mathcal{A}$ is an algebra. To show that it is not a $sigma$-algebra I reasoned as follows: fixed a sequence ${x_n}_{ninmathbb{N}}$ of elements of $X$ such that $x_ine x_j$ for $ine j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n={x_n}$ for all $ninmathbb{N}$. At this point I consider the sets $A_2,A_4dots ,A_{2k},dots$ where $kinmathbb{N}$; I observe that $A_{2k}inmathcal{A}$ for all $kinmathbb{N}$, but the union $$bigcup_{k=1}^{+infty}A_{2k}=bigg{x_2,x_4,dots,x_{2k},dotsbigg}$$
is infinite. On the other hand
begin{equation}
begin{split}
bigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^c=Xsetminusbigcup_{k=1}^{+infty}A_{2k}=& Xcapbigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^{c}=Xcapbigg[bigcup_{k=1}^{+infty}A_{2k+1}bigg]\
=&bigcup_{k=1}^{+infty}A_{2k+1}cuptilde{A},
end{split}
end{equation}

where $tilde{A}$ it is what remains if $X$ is not countable, if $X$ is countable $tilde{A}=emptyset$. Therefore $cup_k A_{2k}$ and $[cup_k A_{2k}]^c$ are both infinite, then $$bigcup_{n=1}^{+infty} A_nnotinmathcal{A}.$$



it's correct? Thanks!










share|cite|improve this question






















  • Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
    – Sorin Tirc
    Nov 18 at 13:52










  • Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
    – Ethan Bolker
    Nov 18 at 13:52













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $X$ be a not empty set.



I must prove that the family $mathcal{A}$ defined as follows
$$mathcal{A}=bigg{Ainmathcal{P}(X);|;A;text{is finite or $A^c$ is finite}bigg}$$ it is not a $sigma$-algebra if $X$ is infinite.



Obviously $mathcal{A}$ is an algebra. To show that it is not a $sigma$-algebra I reasoned as follows: fixed a sequence ${x_n}_{ninmathbb{N}}$ of elements of $X$ such that $x_ine x_j$ for $ine j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n={x_n}$ for all $ninmathbb{N}$. At this point I consider the sets $A_2,A_4dots ,A_{2k},dots$ where $kinmathbb{N}$; I observe that $A_{2k}inmathcal{A}$ for all $kinmathbb{N}$, but the union $$bigcup_{k=1}^{+infty}A_{2k}=bigg{x_2,x_4,dots,x_{2k},dotsbigg}$$
is infinite. On the other hand
begin{equation}
begin{split}
bigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^c=Xsetminusbigcup_{k=1}^{+infty}A_{2k}=& Xcapbigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^{c}=Xcapbigg[bigcup_{k=1}^{+infty}A_{2k+1}bigg]\
=&bigcup_{k=1}^{+infty}A_{2k+1}cuptilde{A},
end{split}
end{equation}

where $tilde{A}$ it is what remains if $X$ is not countable, if $X$ is countable $tilde{A}=emptyset$. Therefore $cup_k A_{2k}$ and $[cup_k A_{2k}]^c$ are both infinite, then $$bigcup_{n=1}^{+infty} A_nnotinmathcal{A}.$$



it's correct? Thanks!










share|cite|improve this question













Let $X$ be a not empty set.



I must prove that the family $mathcal{A}$ defined as follows
$$mathcal{A}=bigg{Ainmathcal{P}(X);|;A;text{is finite or $A^c$ is finite}bigg}$$ it is not a $sigma$-algebra if $X$ is infinite.



Obviously $mathcal{A}$ is an algebra. To show that it is not a $sigma$-algebra I reasoned as follows: fixed a sequence ${x_n}_{ninmathbb{N}}$ of elements of $X$ such that $x_ine x_j$ for $ine j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n={x_n}$ for all $ninmathbb{N}$. At this point I consider the sets $A_2,A_4dots ,A_{2k},dots$ where $kinmathbb{N}$; I observe that $A_{2k}inmathcal{A}$ for all $kinmathbb{N}$, but the union $$bigcup_{k=1}^{+infty}A_{2k}=bigg{x_2,x_4,dots,x_{2k},dotsbigg}$$
is infinite. On the other hand
begin{equation}
begin{split}
bigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^c=Xsetminusbigcup_{k=1}^{+infty}A_{2k}=& Xcapbigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^{c}=Xcapbigg[bigcup_{k=1}^{+infty}A_{2k+1}bigg]\
=&bigcup_{k=1}^{+infty}A_{2k+1}cuptilde{A},
end{split}
end{equation}

where $tilde{A}$ it is what remains if $X$ is not countable, if $X$ is countable $tilde{A}=emptyset$. Therefore $cup_k A_{2k}$ and $[cup_k A_{2k}]^c$ are both infinite, then $$bigcup_{n=1}^{+infty} A_nnotinmathcal{A}.$$



it's correct? Thanks!







proof-verification proof-writing proof-explanation






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asked Nov 18 at 13:48









Jack J.

4571317




4571317












  • Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
    – Sorin Tirc
    Nov 18 at 13:52










  • Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
    – Ethan Bolker
    Nov 18 at 13:52


















  • Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
    – Sorin Tirc
    Nov 18 at 13:52










  • Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
    – Ethan Bolker
    Nov 18 at 13:52
















Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
– Sorin Tirc
Nov 18 at 13:52




Your solution is perfectly fine, you are using that a countable collection of sets in a sigma algebra stays within the sigma algebra, which is part of the definition of a sigma algebra and you are right in proving that this doesn’t hold here
– Sorin Tirc
Nov 18 at 13:52












Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
– Ethan Bolker
Nov 18 at 13:52




Your argument is essentially correct. You struggle a bit more than necessary showing the complement of the union you found is infinite.
– Ethan Bolker
Nov 18 at 13:52










1 Answer
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oldest

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up vote
2
down vote



accepted










Your proof is correct.



A bit shorter:



Let $X$ be infinite.



Then we can write it as $X=Ycup Z$ where:





  • $Y$ and $Z$ are disjoint.


  • $Y$ is countably infinite.


  • $Z$ is infinite.


Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ where the RHS is a countable union of sets that are elements of $mathcal A$.



So $mathcal A$ is not closed under countable unions, hence is not a $sigma$-algebra.






share|cite|improve this answer























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    1 Answer
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    1 Answer
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    active

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    active

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    up vote
    2
    down vote



    accepted










    Your proof is correct.



    A bit shorter:



    Let $X$ be infinite.



    Then we can write it as $X=Ycup Z$ where:





    • $Y$ and $Z$ are disjoint.


    • $Y$ is countably infinite.


    • $Z$ is infinite.


    Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ where the RHS is a countable union of sets that are elements of $mathcal A$.



    So $mathcal A$ is not closed under countable unions, hence is not a $sigma$-algebra.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      Your proof is correct.



      A bit shorter:



      Let $X$ be infinite.



      Then we can write it as $X=Ycup Z$ where:





      • $Y$ and $Z$ are disjoint.


      • $Y$ is countably infinite.


      • $Z$ is infinite.


      Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ where the RHS is a countable union of sets that are elements of $mathcal A$.



      So $mathcal A$ is not closed under countable unions, hence is not a $sigma$-algebra.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Your proof is correct.



        A bit shorter:



        Let $X$ be infinite.



        Then we can write it as $X=Ycup Z$ where:





        • $Y$ and $Z$ are disjoint.


        • $Y$ is countably infinite.


        • $Z$ is infinite.


        Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ where the RHS is a countable union of sets that are elements of $mathcal A$.



        So $mathcal A$ is not closed under countable unions, hence is not a $sigma$-algebra.






        share|cite|improve this answer














        Your proof is correct.



        A bit shorter:



        Let $X$ be infinite.



        Then we can write it as $X=Ycup Z$ where:





        • $Y$ and $Z$ are disjoint.


        • $Y$ is countably infinite.


        • $Z$ is infinite.


        Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ where the RHS is a countable union of sets that are elements of $mathcal A$.



        So $mathcal A$ is not closed under countable unions, hence is not a $sigma$-algebra.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 18 at 20:37

























        answered Nov 18 at 13:59









        drhab

        95.1k543126




        95.1k543126






























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