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-2

$begingroup$

I have this question from a homework set I don’t quite understand the function definition to start on this problem

Problem 5. For this problem, we recall: if f : A → B is a bijection between finite sets A and B, then |A| = |B| namely A and B have the same cardinality (see what is called in your book the pigeonhole principle
For any set X , denote by 0, 1X the set of all functions X → 0, 1. That is, 0,1X = f : f is a function from X to 0,1.
1. Define sets Xn = 1,...,n for all n ∈ N. Find a bijection 0,1Xn+1 −→0,1×0,1Xn.
2. Show by induction on n ∈ N that 0, 1Xn has cardinality 2n.
3. Let X be any set. For each subset Y ⊆ X, define fY to be the function
fY : X −→ 0,1
x ——> { 0 If X in Y
{ 1. If X in XY
Prove that the function
F : Power set (X) → 0,1^X
A → fA
is a bijection.
4. Prove that Power set (Xn) has cardinality 2n.

Really need help with this!

formal-proofs constructive-mathematics

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edited Mar 15 at 6:09

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closed as off-topic by Eevee Trainer, Kavi Rama Murthy, Lord Shark the Unknown, Clayton, Parcly Taxel Mar 16 at 2:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kavi Rama Murthy, Lord Shark the Unknown, Clayton, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
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  $begingroup$
  Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
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  – Eevee Trainer
  Mar 15 at 5:53
  

  
  
  
  

add a comment | 

-2

$begingroup$

I have this question from a homework set I don’t quite understand the function definition to start on this problem

Problem 5. For this problem, we recall: if f : A → B is a bijection between finite sets A and B, then |A| = |B| namely A and B have the same cardinality (see what is called in your book the pigeonhole principle
For any set X , denote by 0, 1X the set of all functions X → 0, 1. That is, 0,1X = f : f is a function from X to 0,1.
1. Define sets Xn = 1,...,n for all n ∈ N. Find a bijection 0,1Xn+1 −→0,1×0,1Xn.
2. Show by induction on n ∈ N that 0, 1Xn has cardinality 2n.
3. Let X be any set. For each subset Y ⊆ X, define fY to be the function
fY : X −→ 0,1
x ——> { 0 If X in Y
{ 1. If X in XY
Prove that the function
F : Power set (X) → 0,1^X
A → fA
is a bijection.
4. Prove that Power set (Xn) has cardinality 2n.

Really need help with this!

formal-proofs constructive-mathematics

share|cite|improve this question

edited Mar 15 at 6:09

Rita

asked Mar 15 at 5:50

RitaRita

112

$endgroup$

closed as off-topic by Eevee Trainer, Kavi Rama Murthy, Lord Shark the Unknown, Clayton, Parcly Taxel Mar 16 at 2:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kavi Rama Murthy, Lord Shark the Unknown, Clayton, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
  $endgroup$
  – Eevee Trainer
  Mar 15 at 5:53
  

  
  
  
  

add a comment | 

$begingroup$

I have this question from a homework set I don’t quite understand the function definition to start on this problem

Problem 5. For this problem, we recall: if f : A → B is a bijection between finite sets A and B, then |A| = |B| namely A and B have the same cardinality (see what is called in your book the pigeonhole principle
For any set X , denote by 0, 1X the set of all functions X → 0, 1. That is, 0,1X = f : f is a function from X to 0,1.
1. Define sets Xn = 1,...,n for all n ∈ N. Find a bijection 0,1Xn+1 −→0,1×0,1Xn.
2. Show by induction on n ∈ N that 0, 1Xn has cardinality 2n.
3. Let X be any set. For each subset Y ⊆ X, define fY to be the function
fY : X −→ 0,1
x ——> { 0 If X in Y
{ 1. If X in XY
Prove that the function
F : Power set (X) → 0,1^X
A → fA
is a bijection.
4. Prove that Power set (Xn) has cardinality 2n.

Really need help with this!

formal-proofs constructive-mathematics

share|cite|improve this question

edited Mar 15 at 6:09

Rita

asked Mar 15 at 5:50

RitaRita

112

$endgroup$

share|cite|improve this question

edited Mar 15 at 6:09

Rita

asked Mar 15 at 5:50

RitaRita

112

share|cite|improve this question

edited Mar 15 at 6:09

Rita

asked Mar 15 at 5:50

RitaRita

112

asked Mar 15 at 5:50

RitaRita

112

asked Mar 15 at 5:50

RitaRita

112

1 Answer
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$begingroup$

This is only a partial answer.

---

For two sets $A,B$ the set of functions from $Ato B$ is a set which contains all possibile different functions that can be created from one set to the other.

Let's for example $A=1,2,3,4$ and look at $0,1A$ be the set of functions from $Ato0,1$. Each element of this set is created by defining the image for each of the four elements of $A$, so the function is defined by four choices:
$f(1), f(2), f(3)$ and $f(4)$, and each of these must be an element of $0,1$ so either $1$ or $2$.

Consequently how many different functions can you construct? It's simple: you have two choices for $f(1)$ two for $f(2)$ and so on. In general the number of possibile functions is nothing but $2^A$.

If the codomain $B$ were to have more than $2$ elements, it's simple to generalize the above argument and get that the set of functions from $Ato B$ has cardinality $|B|^A$.

---

Let's now attack the first question: finding a bijection from $$0,1X_n+1to0,1times0,1X_n.$$
This is actually rather easy:

Just take a function which is defined on $0,dots,n,n+1to 0,1$, restrict it to the function which is defined on $0,dots,nto 0,1$ (i.e. an element of $0,1X_n$)and keep $f(n+1)$ separate (i.e. an element of $0,1$). The looked for bijection is thus:
$$fmapsto(f(n+1),fmid_X_n).$$

I'll leave the others to you for now.

share|cite|improve this answer

edited Mar 15 at 6:31

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

$endgroup$

add a comment | 

0

$begingroup$

This is only a partial answer.

---

For two sets $A,B$ the set of functions from $Ato B$ is a set which contains all possibile different functions that can be created from one set to the other.

Let's for example $A=1,2,3,4$ and look at $0,1A$ be the set of functions from $Ato0,1$. Each element of this set is created by defining the image for each of the four elements of $A$, so the function is defined by four choices:
$f(1), f(2), f(3)$ and $f(4)$, and each of these must be an element of $0,1$ so either $1$ or $2$.

Consequently how many different functions can you construct? It's simple: you have two choices for $f(1)$ two for $f(2)$ and so on. In general the number of possibile functions is nothing but $2^A$.

If the codomain $B$ were to have more than $2$ elements, it's simple to generalize the above argument and get that the set of functions from $Ato B$ has cardinality $|B|^A$.

---

Let's now attack the first question: finding a bijection from $$0,1X_n+1to0,1times0,1X_n.$$
This is actually rather easy:

Just take a function which is defined on $0,dots,n,n+1to 0,1$, restrict it to the function which is defined on $0,dots,nto 0,1$ (i.e. an element of $0,1X_n$)and keep $f(n+1)$ separate (i.e. an element of $0,1$). The looked for bijection is thus:
$$fmapsto(f(n+1),fmid_X_n).$$

I'll leave the others to you for now.

share|cite|improve this answer

edited Mar 15 at 6:31

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

$endgroup$

add a comment | 

0

$begingroup$

This is only a partial answer.

---

For two sets $A,B$ the set of functions from $Ato B$ is a set which contains all possibile different functions that can be created from one set to the other.

Let's for example $A=1,2,3,4$ and look at $0,1A$ be the set of functions from $Ato0,1$. Each element of this set is created by defining the image for each of the four elements of $A$, so the function is defined by four choices:
$f(1), f(2), f(3)$ and $f(4)$, and each of these must be an element of $0,1$ so either $1$ or $2$.

Consequently how many different functions can you construct? It's simple: you have two choices for $f(1)$ two for $f(2)$ and so on. In general the number of possibile functions is nothing but $2^A$.

If the codomain $B$ were to have more than $2$ elements, it's simple to generalize the above argument and get that the set of functions from $Ato B$ has cardinality $|B|^A$.

---

Let's now attack the first question: finding a bijection from $$0,1X_n+1to0,1times0,1X_n.$$
This is actually rather easy:

Just take a function which is defined on $0,dots,n,n+1to 0,1$, restrict it to the function which is defined on $0,dots,nto 0,1$ (i.e. an element of $0,1X_n$)and keep $f(n+1)$ separate (i.e. an element of $0,1$). The looked for bijection is thus:
$$fmapsto(f(n+1),fmid_X_n).$$

I'll leave the others to you for now.

share|cite|improve this answer

edited Mar 15 at 6:31

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

$endgroup$

add a comment | 

$begingroup$

This is only a partial answer.

---

For two sets $A,B$ the set of functions from $Ato B$ is a set which contains all possibile different functions that can be created from one set to the other.

Let's for example $A=1,2,3,4$ and look at $0,1A$ be the set of functions from $Ato0,1$. Each element of this set is created by defining the image for each of the four elements of $A$, so the function is defined by four choices:
$f(1), f(2), f(3)$ and $f(4)$, and each of these must be an element of $0,1$ so either $1$ or $2$.

Consequently how many different functions can you construct? It's simple: you have two choices for $f(1)$ two for $f(2)$ and so on. In general the number of possibile functions is nothing but $2^A$.

If the codomain $B$ were to have more than $2$ elements, it's simple to generalize the above argument and get that the set of functions from $Ato B$ has cardinality $|B|^A$.

---

Let's now attack the first question: finding a bijection from $$0,1X_n+1to0,1times0,1X_n.$$
This is actually rather easy:

Just take a function which is defined on $0,dots,n,n+1to 0,1$, restrict it to the function which is defined on $0,dots,nto 0,1$ (i.e. an element of $0,1X_n$)and keep $f(n+1)$ separate (i.e. an element of $0,1$). The looked for bijection is thus:
$$fmapsto(f(n+1),fmid_X_n).$$

I'll leave the others to you for now.

share|cite|improve this answer

edited Mar 15 at 6:31

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

$endgroup$

share|cite|improve this answer

edited Mar 15 at 6:31

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335

answered Mar 15 at 6:25

b00n heTb00n heT

10.5k12335