Let function $f$ be defined by $f(X)$. Prove that $f$ is bijective.
$begingroup$
Let function $f: mathcal{P}(mathbb{N}times{0,1}) rightarrow mathcal{P}(mathbb{N})$ be defined by $f(X) = {2x+y:|:langle x,yrangle in X}$.
Prove that $f$ is bijective.
$mathcal{P}(X)$ is a power set over $X$.
In order for function to be bijective, it has to be:
- 1-1: $f(langle x_1,y_1rangle) = f(langle x_2,y_2rangle) Rightarrow langle x_1,y_1rangle = langle x_2,y_2rangle$
- onto: $forall z in mathcal{P}(mathbb{N}) :exists x,yinmathcal{P}(mathbb{N}times{0,1}): f(x,y) = z$
To show that $f$ is 1-1, we take any $x_1, x_2in mathcal{P}(mathbb{N})$; $y_1, y_2in {0,1}$ and we need to show that the first condition is true. Similarly, we need to show that the second condition is true, then it would mean that $f$ is bijective, though I do not know how to prove it step-by-step.
I also thought about other possible way of solving it. Function $h:Arightarrow B$ is bijective iff it has an inverse function $h^{-1}:Brightarrow A$, so I could think of a function which would satisfy this condition, but would it be a sufficient proof?
functions elementary-set-theory
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
$endgroup$
add a comment |
$begingroup$
Let function $f: mathcal{P}(mathbb{N}times{0,1}) rightarrow mathcal{P}(mathbb{N})$ be defined by $f(X) = {2x+y:|:langle x,yrangle in X}$.
Prove that $f$ is bijective.
$mathcal{P}(X)$ is a power set over $X$.
In order for function to be bijective, it has to be:
- 1-1: $f(langle x_1,y_1rangle) = f(langle x_2,y_2rangle) Rightarrow langle x_1,y_1rangle = langle x_2,y_2rangle$
- onto: $forall z in mathcal{P}(mathbb{N}) :exists x,yinmathcal{P}(mathbb{N}times{0,1}): f(x,y) = z$
To show that $f$ is 1-1, we take any $x_1, x_2in mathcal{P}(mathbb{N})$; $y_1, y_2in {0,1}$ and we need to show that the first condition is true. Similarly, we need to show that the second condition is true, then it would mean that $f$ is bijective, though I do not know how to prove it step-by-step.
I also thought about other possible way of solving it. Function $h:Arightarrow B$ is bijective iff it has an inverse function $h^{-1}:Brightarrow A$, so I could think of a function which would satisfy this condition, but would it be a sufficient proof?
functions elementary-set-theory
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
$endgroup$
$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47
add a comment |
$begingroup$
Let function $f: mathcal{P}(mathbb{N}times{0,1}) rightarrow mathcal{P}(mathbb{N})$ be defined by $f(X) = {2x+y:|:langle x,yrangle in X}$.
Prove that $f$ is bijective.
$mathcal{P}(X)$ is a power set over $X$.
In order for function to be bijective, it has to be:
- 1-1: $f(langle x_1,y_1rangle) = f(langle x_2,y_2rangle) Rightarrow langle x_1,y_1rangle = langle x_2,y_2rangle$
- onto: $forall z in mathcal{P}(mathbb{N}) :exists x,yinmathcal{P}(mathbb{N}times{0,1}): f(x,y) = z$
To show that $f$ is 1-1, we take any $x_1, x_2in mathcal{P}(mathbb{N})$; $y_1, y_2in {0,1}$ and we need to show that the first condition is true. Similarly, we need to show that the second condition is true, then it would mean that $f$ is bijective, though I do not know how to prove it step-by-step.
I also thought about other possible way of solving it. Function $h:Arightarrow B$ is bijective iff it has an inverse function $h^{-1}:Brightarrow A$, so I could think of a function which would satisfy this condition, but would it be a sufficient proof?
functions elementary-set-theory
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
$endgroup$
Let function $f: mathcal{P}(mathbb{N}times{0,1}) rightarrow mathcal{P}(mathbb{N})$ be defined by $f(X) = {2x+y:|:langle x,yrangle in X}$.
Prove that $f$ is bijective.
$mathcal{P}(X)$ is a power set over $X$.
In order for function to be bijective, it has to be:
- 1-1: $f(langle x_1,y_1rangle) = f(langle x_2,y_2rangle) Rightarrow langle x_1,y_1rangle = langle x_2,y_2rangle$
- onto: $forall z in mathcal{P}(mathbb{N}) :exists x,yinmathcal{P}(mathbb{N}times{0,1}): f(x,y) = z$
To show that $f$ is 1-1, we take any $x_1, x_2in mathcal{P}(mathbb{N})$; $y_1, y_2in {0,1}$ and we need to show that the first condition is true. Similarly, we need to show that the second condition is true, then it would mean that $f$ is bijective, though I do not know how to prove it step-by-step.
I also thought about other possible way of solving it. Function $h:Arightarrow B$ is bijective iff it has an inverse function $h^{-1}:Brightarrow A$, so I could think of a function which would satisfy this condition, but would it be a sufficient proof?
functions elementary-set-theory
functions elementary-set-theory
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 17:10
whiskeyowhiskeyo
1368
1368
$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47
add a comment |
$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47
$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47
add a comment |
1 Answer
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$begingroup$
Hint: Each $Y in mathcal{P}(mathbb{N})$ has a unique decomposition $Y = Y_0 cup Y_1$ where $Y_0$ contains all even numbers in $Y$ and $Y_1$ contains all odd numbers in $Y$. Define
$$Y_0' = { n/2 mid n in Y_0 } ,$$
$$Y_1' = { (n-1)/2 mid n in Y_1 } ,$$
$$g : mathcal{P}(mathbb{N}) to mathcal{P}(mathbb{N} times { 0,1 }), g(Y) = Y_0' times { 0 } cup Y_1' times { 1 } $$
and show that $g$ is an inverse for $f$.
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
$endgroup$
add a comment |
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$begingroup$
Hint: Each $Y in mathcal{P}(mathbb{N})$ has a unique decomposition $Y = Y_0 cup Y_1$ where $Y_0$ contains all even numbers in $Y$ and $Y_1$ contains all odd numbers in $Y$. Define
$$Y_0' = { n/2 mid n in Y_0 } ,$$
$$Y_1' = { (n-1)/2 mid n in Y_1 } ,$$
$$g : mathcal{P}(mathbb{N}) to mathcal{P}(mathbb{N} times { 0,1 }), g(Y) = Y_0' times { 0 } cup Y_1' times { 1 } $$
and show that $g$ is an inverse for $f$.
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
$endgroup$
add a comment |
$begingroup$
Hint: Each $Y in mathcal{P}(mathbb{N})$ has a unique decomposition $Y = Y_0 cup Y_1$ where $Y_0$ contains all even numbers in $Y$ and $Y_1$ contains all odd numbers in $Y$. Define
$$Y_0' = { n/2 mid n in Y_0 } ,$$
$$Y_1' = { (n-1)/2 mid n in Y_1 } ,$$
$$g : mathcal{P}(mathbb{N}) to mathcal{P}(mathbb{N} times { 0,1 }), g(Y) = Y_0' times { 0 } cup Y_1' times { 1 } $$
and show that $g$ is an inverse for $f$.
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
$endgroup$
add a comment |
$begingroup$
Hint: Each $Y in mathcal{P}(mathbb{N})$ has a unique decomposition $Y = Y_0 cup Y_1$ where $Y_0$ contains all even numbers in $Y$ and $Y_1$ contains all odd numbers in $Y$. Define
$$Y_0' = { n/2 mid n in Y_0 } ,$$
$$Y_1' = { (n-1)/2 mid n in Y_1 } ,$$
$$g : mathcal{P}(mathbb{N}) to mathcal{P}(mathbb{N} times { 0,1 }), g(Y) = Y_0' times { 0 } cup Y_1' times { 1 } $$
and show that $g$ is an inverse for $f$.
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
$endgroup$
Hint: Each $Y in mathcal{P}(mathbb{N})$ has a unique decomposition $Y = Y_0 cup Y_1$ where $Y_0$ contains all even numbers in $Y$ and $Y_1$ contains all odd numbers in $Y$. Define
$$Y_0' = { n/2 mid n in Y_0 } ,$$
$$Y_1' = { (n-1)/2 mid n in Y_1 } ,$$
$$g : mathcal{P}(mathbb{N}) to mathcal{P}(mathbb{N} times { 0,1 }), g(Y) = Y_0' times { 0 } cup Y_1' times { 1 } $$
and show that $g$ is an inverse for $f$.
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
edited Dec 13 '18 at 19:19
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
answered Dec 13 '18 at 18:53
Paul FrostPaul Frost
10.5k3933
10.5k3933
add a comment |
add a comment |
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$begingroup$
Yes, both approaches work. But you have to do it. It is not that difficult.
$endgroup$
– Paul Frost
Dec 13 '18 at 17:53
$begingroup$
This is why I asked for help, I mentioned that I cannot prove it step-by-step using first method, as I do not know how should I start it and what I have to do next (sadly proofs are not my strong point). When it comes to the second method with finding the inverse function, I am trying to get it, but all my ideas were wrong. I was thinking about $f^{-1}(n) = n$ for $nin N$, but it does not contain whole ${0,1}$ set, as it is $f^{-1}(n) = n + 0$ in every case.
$endgroup$
– whiskeyo
Dec 13 '18 at 18:47