Does a function that maps from (almost) any natural number to its set of prime factors is surjective?












0












$begingroup$


Let $A$ be the set of all prime numbers, let $B = mathbb{N} - {0,1}$, and let $f : B to P(A)$ be the function that maps any $b in B$ its set of prime factors. e.g, $f(70)={2,5,7}$.



Is $f$ surjective over $P(A)$?



What I think: Yes, it's surjective, because we can take any prime-factors subset in $P(A)$, and the product of its members is a natural number.



However, I'm not quite sure, since $A$ itself is in $P(A)$, and $A$ is an infinite set, and I don't really know if we can represent some $x in B$ with some "infinite quality" as well...Honestly, using Infinity is currently out of my knowledge (or this homework question).



I'd be glad to understand why is itnot surjective, and should I take into account the "infinite quality" of $A$ orand $B$.






functions elementary-set-theory prime-factorization







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$endgroup$












  • $begingroup$
    I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:05












  • $begingroup$
    @ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
    $endgroup$
    – HeyJude
    Nov 24 '18 at 22:07






  • 1




    $begingroup$
    If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:10


















0












$begingroup$


Let $A$ be the set of all prime numbers, let $B = mathbb{N} - {0,1}$, and let $f : B to P(A)$ be the function that maps any $b in B$ its set of prime factors. e.g, $f(70)={2,5,7}$.



Is $f$ surjective over $P(A)$?



What I think: Yes, it's surjective, because we can take any prime-factors subset in $P(A)$, and the product of its members is a natural number.



However, I'm not quite sure, since $A$ itself is in $P(A)$, and $A$ is an infinite set, and I don't really know if we can represent some $x in B$ with some "infinite quality" as well...Honestly, using Infinity is currently out of my knowledge (or this homework question).



I'd be glad to understand why is itnot surjective, and should I take into account the "infinite quality" of $A$ orand $B$.






functions elementary-set-theory prime-factorization







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:05












  • $begingroup$
    @ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
    $endgroup$
    – HeyJude
    Nov 24 '18 at 22:07






  • 1




    $begingroup$
    If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:10
















0












0








0





$begingroup$


Let $A$ be the set of all prime numbers, let $B = mathbb{N} - {0,1}$, and let $f : B to P(A)$ be the function that maps any $b in B$ its set of prime factors. e.g, $f(70)={2,5,7}$.



Is $f$ surjective over $P(A)$?



What I think: Yes, it's surjective, because we can take any prime-factors subset in $P(A)$, and the product of its members is a natural number.



However, I'm not quite sure, since $A$ itself is in $P(A)$, and $A$ is an infinite set, and I don't really know if we can represent some $x in B$ with some "infinite quality" as well...Honestly, using Infinity is currently out of my knowledge (or this homework question).



I'd be glad to understand why is itnot surjective, and should I take into account the "infinite quality" of $A$ orand $B$.






functions elementary-set-theory prime-factorization







share|cite|improve this question









$endgroup$




Let $A$ be the set of all prime numbers, let $B = mathbb{N} - {0,1}$, and let $f : B to P(A)$ be the function that maps any $b in B$ its set of prime factors. e.g, $f(70)={2,5,7}$.



Is $f$ surjective over $P(A)$?



What I think: Yes, it's surjective, because we can take any prime-factors subset in $P(A)$, and the product of its members is a natural number.



However, I'm not quite sure, since $A$ itself is in $P(A)$, and $A$ is an infinite set, and I don't really know if we can represent some $x in B$ with some "infinite quality" as well...Honestly, using Infinity is currently out of my knowledge (or this homework question).



I'd be glad to understand why is itnot surjective, and should I take into account the "infinite quality" of $A$ orand $B$.






functions elementary-set-theory prime-factorization




functions elementary-set-theory prime-factorization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 21:59









HeyJudeHeyJude

1687




1687












  • $begingroup$
    I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:05












  • $begingroup$
    @ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
    $endgroup$
    – HeyJude
    Nov 24 '18 at 22:07






  • 1




    $begingroup$
    If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:10




















  • $begingroup$
    I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:05












  • $begingroup$
    @ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
    $endgroup$
    – HeyJude
    Nov 24 '18 at 22:07






  • 1




    $begingroup$
    If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 22:10


















$begingroup$
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
$endgroup$
– Shubham Johri
Nov 24 '18 at 22:05






$begingroup$
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
$endgroup$
– Shubham Johri
Nov 24 '18 at 22:05














$begingroup$
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
$endgroup$
– HeyJude
Nov 24 '18 at 22:07




$begingroup$
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
$endgroup$
– HeyJude
Nov 24 '18 at 22:07




1




1




$begingroup$
If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
$endgroup$
– Shubham Johri
Nov 24 '18 at 22:10






$begingroup$
If $phi$ is included in the codomain, there is no $bin B$ such that $f(b)={}$. The answer is the same, but it avoids your real concern.
$endgroup$
– Shubham Johri
Nov 24 '18 at 22:10












2 Answers
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$begingroup$

Your argument is fine: the map $f$ is not surjective since, for each $nin B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Your concern is exactly right. If you consider the subset $A$ of $A$, i.e.,
    $$
    A = {2, 3, 5, ldots }
    $$
    then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      active

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      $begingroup$

      Your argument is fine: the map $f$ is not surjective since, for each $nin B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Your argument is fine: the map $f$ is not surjective since, for each $nin B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Your argument is fine: the map $f$ is not surjective since, for each $nin B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.






          share|cite|improve this answer









          $endgroup$



          Your argument is fine: the map $f$ is not surjective since, for each $nin B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 22:02









          José Carlos SantosJosé Carlos Santos

          154k22123226




          154k22123226























              2












              $begingroup$

              Your concern is exactly right. If you consider the subset $A$ of $A$, i.e.,
              $$
              A = {2, 3, 5, ldots }
              $$
              then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your concern is exactly right. If you consider the subset $A$ of $A$, i.e.,
                $$
                A = {2, 3, 5, ldots }
                $$
                then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your concern is exactly right. If you consider the subset $A$ of $A$, i.e.,
                  $$
                  A = {2, 3, 5, ldots }
                  $$
                  then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.






                  share|cite|improve this answer









                  $endgroup$



                  Your concern is exactly right. If you consider the subset $A$ of $A$, i.e.,
                  $$
                  A = {2, 3, 5, ldots }
                  $$
                  then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 '18 at 22:03









                  John HughesJohn Hughes

                  62.7k24090




                  62.7k24090






























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