Prove that union of countable sets is countable












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Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.



enter image description here



Here are the relevant definitions and theorem.



enter image description hereenter image description here






real-analysis proof-explanation







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$endgroup$












  • $begingroup$
    Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 19:05


















1












$begingroup$


Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.



enter image description here



Here are the relevant definitions and theorem.



enter image description hereenter image description here






real-analysis proof-explanation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 19:05
















1












1








1





$begingroup$


Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.



enter image description here



Here are the relevant definitions and theorem.



enter image description hereenter image description here






real-analysis proof-explanation







share|cite|improve this question









$endgroup$




Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.



enter image description here



Here are the relevant definitions and theorem.



enter image description hereenter image description here






real-analysis proof-explanation




real-analysis proof-explanation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 18:55









Josh NgJosh Ng

977




977












  • $begingroup$
    Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 19:05




















  • $begingroup$
    Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 19:05


















$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05






$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05












3 Answers
3






active

oldest

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1












$begingroup$

Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.



      Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
      $$
      T={(x_s,y_s): sin S}.
      $$
      Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
        $endgroup$
        – Martin Argerami
        Dec 12 '18 at 19:15










      • $begingroup$
        Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
        $endgroup$
        – Federico
        Dec 12 '18 at 19:23













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

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      active

      oldest

      votes









      1












      $begingroup$

      Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
      $f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
        $f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
          $f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.






          share|cite|improve this answer









          $endgroup$



          Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
          $f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 19:04









          FedericoFederico

          5,034514




          5,034514























              0












              $begingroup$

              The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.






                  share|cite|improve this answer









                  $endgroup$



                  The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 19:10









                  Joel PereiraJoel Pereira

                  75719




                  75719























                      0












                      $begingroup$

                      What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.



                      Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
                      $$
                      T={(x_s,y_s): sin S}.
                      $$
                      Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                        $endgroup$
                        – Martin Argerami
                        Dec 12 '18 at 19:15










                      • $begingroup$
                        Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                        $endgroup$
                        – Federico
                        Dec 12 '18 at 19:23


















                      0












                      $begingroup$

                      What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.



                      Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
                      $$
                      T={(x_s,y_s): sin S}.
                      $$
                      Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                        $endgroup$
                        – Martin Argerami
                        Dec 12 '18 at 19:15










                      • $begingroup$
                        Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                        $endgroup$
                        – Federico
                        Dec 12 '18 at 19:23
















                      0












                      0








                      0





                      $begingroup$

                      What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.



                      Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
                      $$
                      T={(x_s,y_s): sin S}.
                      $$
                      Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.






                      share|cite|improve this answer











                      $endgroup$



                      What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.



                      Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
                      $$
                      T={(x_s,y_s): sin S}.
                      $$
                      Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 12 '18 at 19:14

























                      answered Dec 12 '18 at 19:06









                      Martin ArgeramiMartin Argerami

                      126k1182181




                      126k1182181












                      • $begingroup$
                        I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                        $endgroup$
                        – Martin Argerami
                        Dec 12 '18 at 19:15










                      • $begingroup$
                        Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                        $endgroup$
                        – Federico
                        Dec 12 '18 at 19:23




















                      • $begingroup$
                        I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                        $endgroup$
                        – Martin Argerami
                        Dec 12 '18 at 19:15










                      • $begingroup$
                        Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                        $endgroup$
                        – Federico
                        Dec 12 '18 at 19:23


















                      $begingroup$
                      I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                      $endgroup$
                      – Martin Argerami
                      Dec 12 '18 at 19:15




                      $begingroup$
                      I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
                      $endgroup$
                      – Martin Argerami
                      Dec 12 '18 at 19:15












                      $begingroup$
                      Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                      $endgroup$
                      – Federico
                      Dec 12 '18 at 19:23






                      $begingroup$
                      Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
                      $endgroup$
                      – Federico
                      Dec 12 '18 at 19:23




















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