Apostol proof for $mathbb{Q}$ being countable.












0












$begingroup$


I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:




The set of rationals $mathbb{Q}$ is countable.




Here is the proof (I rewrote a few things):




Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.




For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?






real-analysis elementary-set-theory proof-explanation rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
    $endgroup$
    – Brennan Vincent
    Jan 7 at 4:07
















0












$begingroup$


I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:




The set of rationals $mathbb{Q}$ is countable.




Here is the proof (I rewrote a few things):




Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.




For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?






real-analysis elementary-set-theory proof-explanation rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
    $endgroup$
    – Brennan Vincent
    Jan 7 at 4:07














0












0








0





$begingroup$


I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:




The set of rationals $mathbb{Q}$ is countable.




Here is the proof (I rewrote a few things):




Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.




For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?






real-analysis elementary-set-theory proof-explanation rational-numbers







share|cite|improve this question











$endgroup$




I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:




The set of rationals $mathbb{Q}$ is countable.




Here is the proof (I rewrote a few things):




Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.




For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?






real-analysis elementary-set-theory proof-explanation rational-numbers




real-analysis elementary-set-theory proof-explanation rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 15:19









Andrés E. Caicedo

65.2k8158247




65.2k8158247










asked Jan 7 at 3:46









numericalorangenumericalorange

1,735311




1,735311












  • $begingroup$
    Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
    $endgroup$
    – Brennan Vincent
    Jan 7 at 4:07


















  • $begingroup$
    Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
    $endgroup$
    – Brennan Vincent
    Jan 7 at 4:07
















$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07




$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07










2 Answers
2






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$begingroup$

Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Rephrasing:



    Fix $i$ , $i in mathbb{Z^+}$.



    $A_i= ${$1/i,2/i,3/i,.......$}.



    $f(mathbb{Z^+}) rightarrow A_i,$



    $f(n)= n/i$, $i$ fixed, is a bijection.



    $A_i$ is countable.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
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      2 Answers
      2






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      4












      $begingroup$

      Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.






          share|cite|improve this answer









          $endgroup$



          Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 3:49









          Tsemo AristideTsemo Aristide

          57.3k11444




          57.3k11444























              2












              $begingroup$

              Rephrasing:



              Fix $i$ , $i in mathbb{Z^+}$.



              $A_i= ${$1/i,2/i,3/i,.......$}.



              $f(mathbb{Z^+}) rightarrow A_i,$



              $f(n)= n/i$, $i$ fixed, is a bijection.



              $A_i$ is countable.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Rephrasing:



                Fix $i$ , $i in mathbb{Z^+}$.



                $A_i= ${$1/i,2/i,3/i,.......$}.



                $f(mathbb{Z^+}) rightarrow A_i,$



                $f(n)= n/i$, $i$ fixed, is a bijection.



                $A_i$ is countable.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Rephrasing:



                  Fix $i$ , $i in mathbb{Z^+}$.



                  $A_i= ${$1/i,2/i,3/i,.......$}.



                  $f(mathbb{Z^+}) rightarrow A_i,$



                  $f(n)= n/i$, $i$ fixed, is a bijection.



                  $A_i$ is countable.






                  share|cite|improve this answer











                  $endgroup$



                  Rephrasing:



                  Fix $i$ , $i in mathbb{Z^+}$.



                  $A_i= ${$1/i,2/i,3/i,.......$}.



                  $f(mathbb{Z^+}) rightarrow A_i,$



                  $f(n)= n/i$, $i$ fixed, is a bijection.



                  $A_i$ is countable.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 7 at 7:19

























                  answered Jan 7 at 6:52









                  Peter SzilasPeter Szilas

                  11.1k2821




                  11.1k2821






























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