Show an infinite sigma algebra contains an infinite sequence of disjoint sets proof clarification












2












$begingroup$


I am new to real analysis and was looking at this problem.
I have been looking at the numerous answers to this question and I feel like I am close to understanding; however, I am not quite understanding this last, bolded part of this Proof (The first answer at the link).



Let $X$ be the whole space. First we show that




there is $Einmathcal{M}$ such that the restriction of $mathcal{M}$ to $E^c$ is still infinite.




If no such $E$ existed, then pick any $emptysetneq Einmathcal{M}$. The restriction of $mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.



Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $sigma$-algebra to $E^c$, $E_2$ from the restriction of the $sigma$-algebra to $E^csetminus E_1$, and so on ...



I think that it is saying let $E_0$ be the first element so that the restriction of $E_0^c$ and $mathcal{M}$ is infinite. But then I don't understand what $E_1$ is.



Can I say that $mathcal{M}_1={F cap E_0^c : Fin mathcal{M}}$. Then we have $mathcal{M}_1$ is still a sigma algebra since it is closed under intersections and compliments. Then Take $E_1$ so that the restriction of $E_1^c$ and $mathcal{M}_1$ is infinite.



I feel like this maybe isn't the right way to do it. I am a bit concerned that I can’t say that the $mathcal{M}_1$ is a sigma algebra.



If someone could explicitly write for me what the $E_i$ are I would really appreciate the clarification. Thank you.






real-analysis measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I feel like it is the way to do it. You should check out the previous posts on the same question.
    $endgroup$
    – Y.Ding
    Aug 22 '18 at 3:10










  • $begingroup$
    That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:20










  • $begingroup$
    Sorry about that. Fixing it now
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:33






  • 1




    $begingroup$
    You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
    $endgroup$
    – Y.Ding
    Aug 23 '18 at 4:50












  • $begingroup$
    @Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
    $endgroup$
    – MathIsHard
    Aug 23 '18 at 14:49
















2












$begingroup$


I am new to real analysis and was looking at this problem.
I have been looking at the numerous answers to this question and I feel like I am close to understanding; however, I am not quite understanding this last, bolded part of this Proof (The first answer at the link).



Let $X$ be the whole space. First we show that




there is $Einmathcal{M}$ such that the restriction of $mathcal{M}$ to $E^c$ is still infinite.




If no such $E$ existed, then pick any $emptysetneq Einmathcal{M}$. The restriction of $mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.



Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $sigma$-algebra to $E^c$, $E_2$ from the restriction of the $sigma$-algebra to $E^csetminus E_1$, and so on ...



I think that it is saying let $E_0$ be the first element so that the restriction of $E_0^c$ and $mathcal{M}$ is infinite. But then I don't understand what $E_1$ is.



Can I say that $mathcal{M}_1={F cap E_0^c : Fin mathcal{M}}$. Then we have $mathcal{M}_1$ is still a sigma algebra since it is closed under intersections and compliments. Then Take $E_1$ so that the restriction of $E_1^c$ and $mathcal{M}_1$ is infinite.



I feel like this maybe isn't the right way to do it. I am a bit concerned that I can’t say that the $mathcal{M}_1$ is a sigma algebra.



If someone could explicitly write for me what the $E_i$ are I would really appreciate the clarification. Thank you.






real-analysis measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I feel like it is the way to do it. You should check out the previous posts on the same question.
    $endgroup$
    – Y.Ding
    Aug 22 '18 at 3:10










  • $begingroup$
    That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:20










  • $begingroup$
    Sorry about that. Fixing it now
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:33






  • 1




    $begingroup$
    You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
    $endgroup$
    – Y.Ding
    Aug 23 '18 at 4:50












  • $begingroup$
    @Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
    $endgroup$
    – MathIsHard
    Aug 23 '18 at 14:49














2












2








2


0



$begingroup$


I am new to real analysis and was looking at this problem.
I have been looking at the numerous answers to this question and I feel like I am close to understanding; however, I am not quite understanding this last, bolded part of this Proof (The first answer at the link).



Let $X$ be the whole space. First we show that




there is $Einmathcal{M}$ such that the restriction of $mathcal{M}$ to $E^c$ is still infinite.




If no such $E$ existed, then pick any $emptysetneq Einmathcal{M}$. The restriction of $mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.



Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $sigma$-algebra to $E^c$, $E_2$ from the restriction of the $sigma$-algebra to $E^csetminus E_1$, and so on ...



I think that it is saying let $E_0$ be the first element so that the restriction of $E_0^c$ and $mathcal{M}$ is infinite. But then I don't understand what $E_1$ is.



Can I say that $mathcal{M}_1={F cap E_0^c : Fin mathcal{M}}$. Then we have $mathcal{M}_1$ is still a sigma algebra since it is closed under intersections and compliments. Then Take $E_1$ so that the restriction of $E_1^c$ and $mathcal{M}_1$ is infinite.



I feel like this maybe isn't the right way to do it. I am a bit concerned that I can’t say that the $mathcal{M}_1$ is a sigma algebra.



If someone could explicitly write for me what the $E_i$ are I would really appreciate the clarification. Thank you.






real-analysis measure-theory







share|cite|improve this question











$endgroup$




I am new to real analysis and was looking at this problem.
I have been looking at the numerous answers to this question and I feel like I am close to understanding; however, I am not quite understanding this last, bolded part of this Proof (The first answer at the link).



Let $X$ be the whole space. First we show that




there is $Einmathcal{M}$ such that the restriction of $mathcal{M}$ to $E^c$ is still infinite.




If no such $E$ existed, then pick any $emptysetneq Einmathcal{M}$. The restriction of $mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.



Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $sigma$-algebra to $E^c$, $E_2$ from the restriction of the $sigma$-algebra to $E^csetminus E_1$, and so on ...



I think that it is saying let $E_0$ be the first element so that the restriction of $E_0^c$ and $mathcal{M}$ is infinite. But then I don't understand what $E_1$ is.



Can I say that $mathcal{M}_1={F cap E_0^c : Fin mathcal{M}}$. Then we have $mathcal{M}_1$ is still a sigma algebra since it is closed under intersections and compliments. Then Take $E_1$ so that the restriction of $E_1^c$ and $mathcal{M}_1$ is infinite.



I feel like this maybe isn't the right way to do it. I am a bit concerned that I can’t say that the $mathcal{M}_1$ is a sigma algebra.



If someone could explicitly write for me what the $E_i$ are I would really appreciate the clarification. Thank you.






real-analysis measure-theory




real-analysis measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 '18 at 14:25







MathIsHard

















asked Aug 22 '18 at 2:28









MathIsHardMathIsHard

1,278516




1,278516








  • 1




    $begingroup$
    I feel like it is the way to do it. You should check out the previous posts on the same question.
    $endgroup$
    – Y.Ding
    Aug 22 '18 at 3:10










  • $begingroup$
    That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:20










  • $begingroup$
    Sorry about that. Fixing it now
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:33






  • 1




    $begingroup$
    You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
    $endgroup$
    – Y.Ding
    Aug 23 '18 at 4:50












  • $begingroup$
    @Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
    $endgroup$
    – MathIsHard
    Aug 23 '18 at 14:49














  • 1




    $begingroup$
    I feel like it is the way to do it. You should check out the previous posts on the same question.
    $endgroup$
    – Y.Ding
    Aug 22 '18 at 3:10










  • $begingroup$
    That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:20










  • $begingroup$
    Sorry about that. Fixing it now
    $endgroup$
    – MathIsHard
    Aug 22 '18 at 3:33






  • 1




    $begingroup$
    You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
    $endgroup$
    – Y.Ding
    Aug 23 '18 at 4:50












  • $begingroup$
    @Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
    $endgroup$
    – MathIsHard
    Aug 23 '18 at 14:49








1




1




$begingroup$
I feel like it is the way to do it. You should check out the previous posts on the same question.
$endgroup$
– Y.Ding
Aug 22 '18 at 3:10




$begingroup$
I feel like it is the way to do it. You should check out the previous posts on the same question.
$endgroup$
– Y.Ding
Aug 22 '18 at 3:10












$begingroup$
That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
$endgroup$
– MathIsHard
Aug 22 '18 at 3:20




$begingroup$
That'd be awesome if I am right :) I was reading all of the ones I could find on MSE and also online. That was the easiest for me to understand that I posted, but I was having trouble with the last part :/
$endgroup$
– MathIsHard
Aug 22 '18 at 3:20












$begingroup$
Sorry about that. Fixing it now
$endgroup$
– MathIsHard
Aug 22 '18 at 3:33




$begingroup$
Sorry about that. Fixing it now
$endgroup$
– MathIsHard
Aug 22 '18 at 3:33




1




1




$begingroup$
You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
$endgroup$
– Y.Ding
Aug 23 '18 at 4:50






$begingroup$
You can visualize this process by drawing circles (Venn diagram). Basically, you can keep drawing non-empty, disjoint circles indefinitely. And the result is infinitely many non-empty, disjoint circles. You can also get some inspiration by thinking about the following question: why an infinite binary tree must have an infinite branch? Finally, for your problem, can you define a $mathcal{M}_2$ by restricting $mathcal{M}_1$ to $E_1^c$? Then you can get $E_2$ from $mathcal{M}_2$. In general, you can get $mathcal{M}_{i_1}$ from $mathcal{M}_i$ and $E_i^c$.
$endgroup$
– Y.Ding
Aug 23 '18 at 4:50














$begingroup$
@Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
$endgroup$
– MathIsHard
Aug 23 '18 at 14:49




$begingroup$
@Y.Ding I think I can do that yes. I also was able to get help showing that the restriction of an element and the sigma algebra generates another sigma algebra so I can repeat my argument I believe infinitely many times :) as for the venn diagram, that is a great visualization, I was wondering how I know that a circle that I draw is actually one of the sets in the sigma algebra? Thank you very much for your help.
$endgroup$
– MathIsHard
Aug 23 '18 at 14:49










1 Answer
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$begingroup$

From the comments above.




Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $mathcal{M}$ such that $mathcal{M}_1 := { F cap E_0^c : F in mathcal{M} }$ is infinite. You can check that $mathcal{M}_1$ is indeed a $sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $mathcal{M}_1$ such that $mathcal{M}_2 := { F_1 cap E_1^c : F_1 in mathcal{M}_1 }$ is infinite. And you can proceed like this inductively to complete the proof.






share|cite|improve this answer











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    $begingroup$

    From the comments above.




    Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $mathcal{M}$ such that $mathcal{M}_1 := { F cap E_0^c : F in mathcal{M} }$ is infinite. You can check that $mathcal{M}_1$ is indeed a $sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $mathcal{M}_1$ such that $mathcal{M}_2 := { F_1 cap E_1^c : F_1 in mathcal{M}_1 }$ is infinite. And you can proceed like this inductively to complete the proof.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      From the comments above.




      Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $mathcal{M}$ such that $mathcal{M}_1 := { F cap E_0^c : F in mathcal{M} }$ is infinite. You can check that $mathcal{M}_1$ is indeed a $sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $mathcal{M}_1$ such that $mathcal{M}_2 := { F_1 cap E_1^c : F_1 in mathcal{M}_1 }$ is infinite. And you can proceed like this inductively to complete the proof.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        From the comments above.




        Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $mathcal{M}$ such that $mathcal{M}_1 := { F cap E_0^c : F in mathcal{M} }$ is infinite. You can check that $mathcal{M}_1$ is indeed a $sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $mathcal{M}_1$ such that $mathcal{M}_2 := { F_1 cap E_1^c : F_1 in mathcal{M}_1 }$ is infinite. And you can proceed like this inductively to complete the proof.






        share|cite|improve this answer











        $endgroup$



        From the comments above.




        Yes, your understanding is correct! You choose $E_0$ to be a nonempty member of $mathcal{M}$ such that $mathcal{M}_1 := { F cap E_0^c : F in mathcal{M} }$ is infinite. You can check that $mathcal{M}_1$ is indeed a $sigma$-algebra, and now you can choose $E_1$ to be a nonempty member of $mathcal{M}_1$ such that $mathcal{M}_2 := { F_1 cap E_1^c : F_1 in mathcal{M}_1 }$ is infinite. And you can proceed like this inductively to complete the proof.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Dec 12 '18 at 13:21


























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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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