Does a function that maps from (almost) any natural number to its set of prime factors is surjective?
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
asked Nov 24 at 21:59
HeyJude
1565
add a comment |
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
asked Nov 24 at 21:59
HeyJude
1565
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
1
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.
– Shubham Johri
Nov 24 at 22:10
add a comment |
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
asked Nov 24 at 21:59
HeyJude
1565
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
asked Nov 24 at 21:59
HeyJude
1565
asked Nov 24 at 21:59
HeyJude
1565
asked Nov 24 at 21:59
HeyJude
1565
asked Nov 24 at 21:59
HeyJude
1565
asked Nov 24 at 21:59
HeyJude
1565
1565
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
1
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.
– Shubham Johri
Nov 24 at 22:10
add a comment |
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
1
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.
– Shubham Johri
Nov 24 at 22:10
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
1
1
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.
– Shubham Johri
Nov 24 at 22:10
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.
– Shubham Johri
Nov 24 at 22:10
add a comment |
2 Answers
2
active
oldest
votes
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 $f$.
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
add a comment |
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 $n$.
answered Nov 24 at 22:03
John Hughes
62.3k24090
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $f$.
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
add a comment |
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 $f$.
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
add a comment |
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 $f$.
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
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 $f$.
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
answered Nov 24 at 22:02
José Carlos Santos
148k22117218
148k22117218
add a comment |
add a comment |
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 $n$.
answered Nov 24 at 22:03
John Hughes
62.3k24090
add a comment |
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 $n$.
answered Nov 24 at 22:03
John Hughes
62.3k24090
add a comment |
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 $n$.
answered Nov 24 at 22:03
John Hughes
62.3k24090
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 $n$.
answered Nov 24 at 22:03
John Hughes
62.3k24090
answered Nov 24 at 22:03
John Hughes
62.3k24090
answered Nov 24 at 22:03
John Hughes
62.3k24090
answered Nov 24 at 22:03
John Hughes
62.3k24090
62.3k24090
add a comment |
add a comment |
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I think you mean to ask whether $f$ is surjective over $P(A)-{phi}$.
– Shubham Johri
Nov 24 at 22:05
@ShubhamJohri, I asked as I intended, but will the answer be different with your rephrasing?
– HeyJude
Nov 24 at 22:07
1
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.
– Shubham Johri
Nov 24 at 22:10