Proof that $sigma$-algebra is not countable (proof revision)












0












$begingroup$


This problem has already several answers here
Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable

and it has been asked a lot times before.


But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
This is the problem:




Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that

(a) $mathcal{M}$ contains an infinite sequence of disjoint sets

(b) $mathcal{M}$ is not countable




So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
$$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.

For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.


Thanks in advance!






real-analysis measure-theory proof-verification







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$endgroup$

















    0












    $begingroup$


    This problem has already several answers here
    Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable

    and it has been asked a lot times before.


    But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
    This is the problem:




    Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that

    (a) $mathcal{M}$ contains an infinite sequence of disjoint sets

    (b) $mathcal{M}$ is not countable




    So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
    $$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
    cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
    so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.

    For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.


    Thanks in advance!






    real-analysis measure-theory proof-verification







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      This problem has already several answers here
      Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable

      and it has been asked a lot times before.


      But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
      This is the problem:




      Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that

      (a) $mathcal{M}$ contains an infinite sequence of disjoint sets

      (b) $mathcal{M}$ is not countable




      So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
      $$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
      cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
      so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.

      For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.


      Thanks in advance!






      real-analysis measure-theory proof-verification







      share|cite|improve this question











      $endgroup$




      This problem has already several answers here
      Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable

      and it has been asked a lot times before.


      But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
      This is the problem:




      Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that

      (a) $mathcal{M}$ contains an infinite sequence of disjoint sets

      (b) $mathcal{M}$ is not countable




      So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
      $$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
      cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
      so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.

      For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.


      Thanks in advance!






      real-analysis measure-theory proof-verification




      real-analysis measure-theory proof-verification






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 18 '18 at 7:37









      Did

      248k23224463




      248k23224463










      asked Dec 18 '18 at 6:37









      mate89mate89

      1799




      1799






















          1 Answer
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          active

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          1












          $begingroup$

          There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
          You will need to do more work to ensure we get non-empty sets.



          The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
            $endgroup$
            – mate89
            Dec 19 '18 at 4:23












          • $begingroup$
            @mate89 the image can be any infinite subset of $omega$
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:11










          • $begingroup$
            Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
            $endgroup$
            – mate89
            Dec 19 '18 at 5:22












          • $begingroup$
            @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:42











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
          You will need to do more work to ensure we get non-empty sets.



          The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
            $endgroup$
            – mate89
            Dec 19 '18 at 4:23












          • $begingroup$
            @mate89 the image can be any infinite subset of $omega$
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:11










          • $begingroup$
            Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
            $endgroup$
            – mate89
            Dec 19 '18 at 5:22












          • $begingroup$
            @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:42
















          1












          $begingroup$

          There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
          You will need to do more work to ensure we get non-empty sets.



          The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
            $endgroup$
            – mate89
            Dec 19 '18 at 4:23












          • $begingroup$
            @mate89 the image can be any infinite subset of $omega$
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:11










          • $begingroup$
            Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
            $endgroup$
            – mate89
            Dec 19 '18 at 5:22












          • $begingroup$
            @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:42














          1












          1








          1





          $begingroup$

          There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
          You will need to do more work to ensure we get non-empty sets.



          The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.






          share|cite|improve this answer









          $endgroup$



          There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
          You will need to do more work to ensure we get non-empty sets.



          The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 7:04









          Henno BrandsmaHenno Brandsma

          110k347117




          110k347117












          • $begingroup$
            Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
            $endgroup$
            – mate89
            Dec 19 '18 at 4:23












          • $begingroup$
            @mate89 the image can be any infinite subset of $omega$
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:11










          • $begingroup$
            Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
            $endgroup$
            – mate89
            Dec 19 '18 at 5:22












          • $begingroup$
            @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:42


















          • $begingroup$
            Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
            $endgroup$
            – mate89
            Dec 19 '18 at 4:23












          • $begingroup$
            @mate89 the image can be any infinite subset of $omega$
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:11










          • $begingroup$
            Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
            $endgroup$
            – mate89
            Dec 19 '18 at 5:22












          • $begingroup$
            @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
            $endgroup$
            – Henno Brandsma
            Dec 19 '18 at 5:42
















          $begingroup$
          Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
          $endgroup$
          – mate89
          Dec 19 '18 at 4:23






          $begingroup$
          Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
          $endgroup$
          – mate89
          Dec 19 '18 at 4:23














          $begingroup$
          @mate89 the image can be any infinite subset of $omega$
          $endgroup$
          – Henno Brandsma
          Dec 19 '18 at 5:11




          $begingroup$
          @mate89 the image can be any infinite subset of $omega$
          $endgroup$
          – Henno Brandsma
          Dec 19 '18 at 5:11












          $begingroup$
          Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
          $endgroup$
          – mate89
          Dec 19 '18 at 5:22






          $begingroup$
          Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
          $endgroup$
          – mate89
          Dec 19 '18 at 5:22














          $begingroup$
          @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
          $endgroup$
          – Henno Brandsma
          Dec 19 '18 at 5:42




          $begingroup$
          @mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
          $endgroup$
          – Henno Brandsma
          Dec 19 '18 at 5:42


















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