Find and prove that a proper subset $Asubset Bbb N$ such that $Aapprox Bbb N$.
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Can anyone check my working please?
Find and prove that a proper subset $Asubset Bbb N$ such that $Aapprox Bbb N$.
Let $A$ be the set of even numbers, it is clear that $Asubset Bbb N$ since all even numbers are a member of $Bbb N$, but $3in Bbb N$ is not in A.
$Aapprox Bbb N$ iff there is a bijection from $A$ to $Bbb N$, ie. a function that is one-one and onto.
Let this function be $f:A→ Bbb N$ such that $f(x)=x÷2$.
One-one: $f(x)=f(y)to x=y$
According to the def. of $f$, this means $frac x2=frac y2to x=y$. But if we times $x$ and $y$ in the antecedent by 2 we get $x=yto x=y$, thus $f$ is one-one.
Onto: $forall nin Bbb N exists xin A$ such that $ f(x)=n$
Assume $ f(x)not=n$, this means there exists an $x$ such that dividing it by 2 does not give us a natural number - but that's impossible. Therefore $f$ is onto.
elementary-set-theory
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
$endgroup$
add a comment |
$begingroup$
Can anyone check my working please?
Find and prove that a proper subset $Asubset Bbb N$ such that $Aapprox Bbb N$.
Let $A$ be the set of even numbers, it is clear that $Asubset Bbb N$ since all even numbers are a member of $Bbb N$, but $3in Bbb N$ is not in A.
$Aapprox Bbb N$ iff there is a bijection from $A$ to $Bbb N$, ie. a function that is one-one and onto.
Let this function be $f:A→ Bbb N$ such that $f(x)=x÷2$.
One-one: $f(x)=f(y)to x=y$
According to the def. of $f$, this means $frac x2=frac y2to x=y$. But if we times $x$ and $y$ in the antecedent by 2 we get $x=yto x=y$, thus $f$ is one-one.
Onto: $forall nin Bbb N exists xin A$ such that $ f(x)=n$
Assume $ f(x)not=n$, this means there exists an $x$ such that dividing it by 2 does not give us a natural number - but that's impossible. Therefore $f$ is onto.
elementary-set-theory
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
$endgroup$
1
$begingroup$
Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
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– pwerth
Jan 23 at 20:21
1
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27
add a comment |
$begingroup$
Can anyone check my working please?
Find and prove that a proper subset $Asubset Bbb N$ such that $Aapprox Bbb N$.
Let $A$ be the set of even numbers, it is clear that $Asubset Bbb N$ since all even numbers are a member of $Bbb N$, but $3in Bbb N$ is not in A.
$Aapprox Bbb N$ iff there is a bijection from $A$ to $Bbb N$, ie. a function that is one-one and onto.
Let this function be $f:A→ Bbb N$ such that $f(x)=x÷2$.
One-one: $f(x)=f(y)to x=y$
According to the def. of $f$, this means $frac x2=frac y2to x=y$. But if we times $x$ and $y$ in the antecedent by 2 we get $x=yto x=y$, thus $f$ is one-one.
Onto: $forall nin Bbb N exists xin A$ such that $ f(x)=n$
Assume $ f(x)not=n$, this means there exists an $x$ such that dividing it by 2 does not give us a natural number - but that's impossible. Therefore $f$ is onto.
elementary-set-theory
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
$endgroup$
Can anyone check my working please?
Find and prove that a proper subset $Asubset Bbb N$ such that $Aapprox Bbb N$.
Let $A$ be the set of even numbers, it is clear that $Asubset Bbb N$ since all even numbers are a member of $Bbb N$, but $3in Bbb N$ is not in A.
$Aapprox Bbb N$ iff there is a bijection from $A$ to $Bbb N$, ie. a function that is one-one and onto.
Let this function be $f:A→ Bbb N$ such that $f(x)=x÷2$.
One-one: $f(x)=f(y)to x=y$
According to the def. of $f$, this means $frac x2=frac y2to x=y$. But if we times $x$ and $y$ in the antecedent by 2 we get $x=yto x=y$, thus $f$ is one-one.
Onto: $forall nin Bbb N exists xin A$ such that $ f(x)=n$
Assume $ f(x)not=n$, this means there exists an $x$ such that dividing it by 2 does not give us a natural number - but that's impossible. Therefore $f$ is onto.
elementary-set-theory
elementary-set-theory
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
edited Jan 23 at 20:18
Daniel Mak
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
asked Jan 23 at 20:14
Daniel MakDaniel Mak
490416
490416
1
$begingroup$
Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
$endgroup$
– pwerth
Jan 23 at 20:21
1
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27
add a comment |
1
$begingroup$
Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
$endgroup$
– pwerth
Jan 23 at 20:21
1
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27
1
1
$begingroup$
Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
$endgroup$
– pwerth
Jan 23 at 20:21
$begingroup$
Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
$endgroup$
– pwerth
Jan 23 at 20:21
1
1
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27
add a comment |
1 Answer
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Your example is a good one, and the injectivity proof is good.
Surjectivity, on the other hand, is a bit off. Given an arbitrary $ninBbb N$, you have to show that there is an $xin A$ such that $f(x)=n$. Setting $x=2n$ works.
answered Jan 23 at 20:22
ArthurArthur
117k7116200
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add a comment |
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$begingroup$
Your example is a good one, and the injectivity proof is good.
Surjectivity, on the other hand, is a bit off. Given an arbitrary $ninBbb N$, you have to show that there is an $xin A$ such that $f(x)=n$. Setting $x=2n$ works.
answered Jan 23 at 20:22
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
Your example is a good one, and the injectivity proof is good.
Surjectivity, on the other hand, is a bit off. Given an arbitrary $ninBbb N$, you have to show that there is an $xin A$ such that $f(x)=n$. Setting $x=2n$ works.
answered Jan 23 at 20:22
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
Your example is a good one, and the injectivity proof is good.
Surjectivity, on the other hand, is a bit off. Given an arbitrary $ninBbb N$, you have to show that there is an $xin A$ such that $f(x)=n$. Setting $x=2n$ works.
answered Jan 23 at 20:22
ArthurArthur
117k7116200
$endgroup$
Your example is a good one, and the injectivity proof is good.
Surjectivity, on the other hand, is a bit off. Given an arbitrary $ninBbb N$, you have to show that there is an $xin A$ such that $f(x)=n$. Setting $x=2n$ works.
answered Jan 23 at 20:22
ArthurArthur
117k7116200
answered Jan 23 at 20:22
ArthurArthur
117k7116200
answered Jan 23 at 20:22
ArthurArthur
117k7116200
answered Jan 23 at 20:22
ArthurArthur
117k7116200
117k7116200
add a comment |
add a comment |
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Your phrasing of the proof of surjectivity is wrong. There are many natural numbers such that dividing them by two does not give a natural number - namely, all odd natural numbers. Simply note that for all $ninmathbb{N}$, the number $x=2n$ satisfies $f(x)=n$. So $f$ is onto
$endgroup$
– pwerth
Jan 23 at 20:21
1
$begingroup$
I would change "if we times $x$ and $y$ ..." to "if we multiply each of $x$ and $y$ ...", because "times" is not a verb. The added "each" makes it clear that we're not saying to compute $xy$.
$endgroup$
– John Hughes
Jan 23 at 20:27