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Prove that $|Atimes B| = |Btimes A|$ for any infinte sets $A,B$

How to prove that the set of rational numbers that are not integers is countable and infinte?How to prove this family of sets is a subset of another family of setsfinding counterexample for identity of setsProving that the sets are equal.Find all the sets $A$ and $B$ such that $P(A times B)=P(A)times P(B)$Let $A,B,C$ be sets. Prove that $(A setminus B)setminus C = A setminus (B cup C)$ for any sets $A,B,C$.How to prove union and intersection of classes if sets?Q: Prove that Cartesian Product is distributive over Minkowski sumsHow can I prove that $f:A_{n+1}times prod_{iin I_n}A_itoprod_{iin I_{n+1}}A_i$ is bijective?Prove that if sets $A$ and $B$ are finite, then $|Atimes B| = |A|cdot|B|$.

0

$begingroup$

Prove that $|Atimes B| = |Btimes A|$ for any infinte sets $A,B$

I tried to build $f:(Atimes B) to Btimes A$ that takes $(a_i,b_i)$ and return the opposite $(b_i,a_i)$, but I'm not sure its right.

discrete-mathematics elementary-set-theory

share|cite|improve this question

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That function does the trick, so yes.
  $endgroup$
  – b00n heT
  Mar 13 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's correct.
  $endgroup$
  – Dbchatto67
  Mar 13 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how can i explain that the function is onto BXA?
  $endgroup$
  – yoav amenou
  Mar 13 at 6:45
  

  
  
  
  

add a comment | 

0

$begingroup$

Prove that $|Atimes B| = |Btimes A|$ for any infinte sets $A,B$

I tried to build $f:(Atimes B) to Btimes A$ that takes $(a_i,b_i)$ and return the opposite $(b_i,a_i)$, but I'm not sure its right.

discrete-mathematics elementary-set-theory

share|cite|improve this question

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That function does the trick, so yes.
  $endgroup$
  – b00n heT
  Mar 13 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's correct.
  $endgroup$
  – Dbchatto67
  Mar 13 at 6:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how can i explain that the function is onto BXA?
  $endgroup$
  – yoav amenou
  Mar 13 at 6:45
  

  
  
  
  

add a comment | 

$begingroup$

Prove that $|Atimes B| = |Btimes A|$ for any infinte sets $A,B$

I tried to build $f:(Atimes B) to Btimes A$ that takes $(a_i,b_i)$ and return the opposite $(b_i,a_i)$, but I'm not sure its right.

discrete-mathematics elementary-set-theory

share|cite|improve this question

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

$endgroup$

share|cite|improve this question

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

share|cite|improve this question

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

edited Mar 13 at 6:57

Yadati Kiran

2,1121622

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

asked Mar 13 at 6:33

yoav amenouyoav amenou

224

2 Answers
2

active

oldest

votes

2

$begingroup$

Your idea is good. Let $f : A times B to B times A$ be defined by $f((a,b)):=(b,a).$

Are you able to show that $f$ is injective ?

$f$ is onto: to this end let $(u,v) in B times A$. Then $(v,u) in A times B$ and $f((v,u))=(u,v).$

share|cite|improve this answer

answered Mar 13 at 6:55

FredFred

48.6k11849

$endgroup$

add a comment | 

0

$begingroup$

You can make a far more general claim, as follows: given a family of sets $A$ indexed by set $I$ and an arbitrary permutation $sigma in Sigma(I)$ (the latter denoting the set of all bijections from $I$ to itself), then one has:

$$prod_{i in I} A_{i} approx prod_{i in I} A_{sigma(i)} (mathrm{Ens})$$

where for arbitrary sets $X, Y$ the relation $X approx Y (mathrm{Ens})$ means the two are isomorphic in $mathrm{Ens}$, the category of sets, in other words there exists a bijection between them.

By the very nature of the cartesian product, one can construct a natural pair of mutually inverse bijections that will establish the stated isomorphism of sets. For arbitrary $i in I$ define $$p_{i}: prod_{j in I} A_{j} to A_{i}$$ as the $i$-projection of the left-hand side cartesian product and $$p'_{i}: prod_{j in I} A_{sigma(j)} to A_{sigma(i)}$$ the analogous projection for the right-hand side cartesian product (the $i$-projection assigns to every element $x$ in the respective cartesian product its $i$-component). Furthermore, let us define maps$$q_{i}=p_{sigma(i)},\ q'_{i}=p'_{sigma^{-1}(i)}$$

By the universality property of cartesian products, there will exist two maps
$$f: prod_{i in I} A_{i} to prod_{i in I} A_{sigma(i)},\ g: prod_{i in I} A_{sigma(i)} to prod_{i in I} A_{i}$$ unique with the respective properties that for every $i in I$ one has $p'_{i} circ f=q_i, p_i circ g=q'_{i}$. From this point on it is easy to verify that:
$$p_i circ g circ f=q'_i circ f=p'_{sigma^{-1}(i)} circ f=q_{sigma^{-1}(i)}=p_i$$ and similarly $$p'_i circ f circ g=q_i circ g=p_{sigma(i)} circ g=q'_{sigma(i)}=p'_i$$ for every index $i$, fact which entails the conclusion that $$f circ g=mathbf{1}_{prod_{i in I} A_{sigma(i)}},\ g circ f=mathbf{1}_{prod_{i in I} A_i}$$

In your particular case, this general result can be applied for a family of sets $C$ indexed by the set ${1, 2}$, such that $C_1=A, C_2=B$ together with the transposition $(12)$ (the only nontrivial permutation of the given two-element set, the one that swaps $1$ with $2$). According to the general argument, you obtain bijection $f: A times B to B times A$ given by the correspondence law $f(x,y)=(y,x)$. It is immediately verified that $f$ is its own inverse (what we call an involution), and that is the most direct way to settle the matter, rather than argue individually about injectivity and surjectivity.

share|cite|improve this answer

edited Mar 13 at 16:30

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

$endgroup$

add a comment | 

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2

$begingroup$

Your idea is good. Let $f : A times B to B times A$ be defined by $f((a,b)):=(b,a).$

Are you able to show that $f$ is injective ?

$f$ is onto: to this end let $(u,v) in B times A$. Then $(v,u) in A times B$ and $f((v,u))=(u,v).$

share|cite|improve this answer

answered Mar 13 at 6:55

FredFred

48.6k11849

$endgroup$

add a comment | 

2

$begingroup$

Your idea is good. Let $f : A times B to B times A$ be defined by $f((a,b)):=(b,a).$

Are you able to show that $f$ is injective ?

$f$ is onto: to this end let $(u,v) in B times A$. Then $(v,u) in A times B$ and $f((v,u))=(u,v).$

share|cite|improve this answer

answered Mar 13 at 6:55

FredFred

48.6k11849

$endgroup$

add a comment | 

$begingroup$

Your idea is good. Let $f : A times B to B times A$ be defined by $f((a,b)):=(b,a).$

Are you able to show that $f$ is injective ?

$f$ is onto: to this end let $(u,v) in B times A$. Then $(v,u) in A times B$ and $f((v,u))=(u,v).$

share|cite|improve this answer

answered Mar 13 at 6:55

FredFred

48.6k11849

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:55

FredFred

48.6k11849

answered Mar 13 at 6:55

FredFred

48.6k11849

answered Mar 13 at 6:55

FredFred

48.6k11849

0

$begingroup$

You can make a far more general claim, as follows: given a family of sets $A$ indexed by set $I$ and an arbitrary permutation $sigma in Sigma(I)$ (the latter denoting the set of all bijections from $I$ to itself), then one has:

$$prod_{i in I} A_{i} approx prod_{i in I} A_{sigma(i)} (mathrm{Ens})$$

where for arbitrary sets $X, Y$ the relation $X approx Y (mathrm{Ens})$ means the two are isomorphic in $mathrm{Ens}$, the category of sets, in other words there exists a bijection between them.

By the very nature of the cartesian product, one can construct a natural pair of mutually inverse bijections that will establish the stated isomorphism of sets. For arbitrary $i in I$ define $$p_{i}: prod_{j in I} A_{j} to A_{i}$$ as the $i$-projection of the left-hand side cartesian product and $$p'_{i}: prod_{j in I} A_{sigma(j)} to A_{sigma(i)}$$ the analogous projection for the right-hand side cartesian product (the $i$-projection assigns to every element $x$ in the respective cartesian product its $i$-component). Furthermore, let us define maps$$q_{i}=p_{sigma(i)},\ q'_{i}=p'_{sigma^{-1}(i)}$$

By the universality property of cartesian products, there will exist two maps
$$f: prod_{i in I} A_{i} to prod_{i in I} A_{sigma(i)},\ g: prod_{i in I} A_{sigma(i)} to prod_{i in I} A_{i}$$ unique with the respective properties that for every $i in I$ one has $p'_{i} circ f=q_i, p_i circ g=q'_{i}$. From this point on it is easy to verify that:
$$p_i circ g circ f=q'_i circ f=p'_{sigma^{-1}(i)} circ f=q_{sigma^{-1}(i)}=p_i$$ and similarly $$p'_i circ f circ g=q_i circ g=p_{sigma(i)} circ g=q'_{sigma(i)}=p'_i$$ for every index $i$, fact which entails the conclusion that $$f circ g=mathbf{1}_{prod_{i in I} A_{sigma(i)}},\ g circ f=mathbf{1}_{prod_{i in I} A_i}$$

In your particular case, this general result can be applied for a family of sets $C$ indexed by the set ${1, 2}$, such that $C_1=A, C_2=B$ together with the transposition $(12)$ (the only nontrivial permutation of the given two-element set, the one that swaps $1$ with $2$). According to the general argument, you obtain bijection $f: A times B to B times A$ given by the correspondence law $f(x,y)=(y,x)$. It is immediately verified that $f$ is its own inverse (what we call an involution), and that is the most direct way to settle the matter, rather than argue individually about injectivity and surjectivity.

share|cite|improve this answer

edited Mar 13 at 16:30

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

$endgroup$

add a comment | 

0

$begingroup$

You can make a far more general claim, as follows: given a family of sets $A$ indexed by set $I$ and an arbitrary permutation $sigma in Sigma(I)$ (the latter denoting the set of all bijections from $I$ to itself), then one has:

$$prod_{i in I} A_{i} approx prod_{i in I} A_{sigma(i)} (mathrm{Ens})$$

where for arbitrary sets $X, Y$ the relation $X approx Y (mathrm{Ens})$ means the two are isomorphic in $mathrm{Ens}$, the category of sets, in other words there exists a bijection between them.

By the very nature of the cartesian product, one can construct a natural pair of mutually inverse bijections that will establish the stated isomorphism of sets. For arbitrary $i in I$ define $$p_{i}: prod_{j in I} A_{j} to A_{i}$$ as the $i$-projection of the left-hand side cartesian product and $$p'_{i}: prod_{j in I} A_{sigma(j)} to A_{sigma(i)}$$ the analogous projection for the right-hand side cartesian product (the $i$-projection assigns to every element $x$ in the respective cartesian product its $i$-component). Furthermore, let us define maps$$q_{i}=p_{sigma(i)},\ q'_{i}=p'_{sigma^{-1}(i)}$$

By the universality property of cartesian products, there will exist two maps
$$f: prod_{i in I} A_{i} to prod_{i in I} A_{sigma(i)},\ g: prod_{i in I} A_{sigma(i)} to prod_{i in I} A_{i}$$ unique with the respective properties that for every $i in I$ one has $p'_{i} circ f=q_i, p_i circ g=q'_{i}$. From this point on it is easy to verify that:
$$p_i circ g circ f=q'_i circ f=p'_{sigma^{-1}(i)} circ f=q_{sigma^{-1}(i)}=p_i$$ and similarly $$p'_i circ f circ g=q_i circ g=p_{sigma(i)} circ g=q'_{sigma(i)}=p'_i$$ for every index $i$, fact which entails the conclusion that $$f circ g=mathbf{1}_{prod_{i in I} A_{sigma(i)}},\ g circ f=mathbf{1}_{prod_{i in I} A_i}$$

In your particular case, this general result can be applied for a family of sets $C$ indexed by the set ${1, 2}$, such that $C_1=A, C_2=B$ together with the transposition $(12)$ (the only nontrivial permutation of the given two-element set, the one that swaps $1$ with $2$). According to the general argument, you obtain bijection $f: A times B to B times A$ given by the correspondence law $f(x,y)=(y,x)$. It is immediately verified that $f$ is its own inverse (what we call an involution), and that is the most direct way to settle the matter, rather than argue individually about injectivity and surjectivity.

share|cite|improve this answer

edited Mar 13 at 16:30

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

$endgroup$

add a comment | 

$begingroup$

You can make a far more general claim, as follows: given a family of sets $A$ indexed by set $I$ and an arbitrary permutation $sigma in Sigma(I)$ (the latter denoting the set of all bijections from $I$ to itself), then one has:

$$prod_{i in I} A_{i} approx prod_{i in I} A_{sigma(i)} (mathrm{Ens})$$

where for arbitrary sets $X, Y$ the relation $X approx Y (mathrm{Ens})$ means the two are isomorphic in $mathrm{Ens}$, the category of sets, in other words there exists a bijection between them.

By the very nature of the cartesian product, one can construct a natural pair of mutually inverse bijections that will establish the stated isomorphism of sets. For arbitrary $i in I$ define $$p_{i}: prod_{j in I} A_{j} to A_{i}$$ as the $i$-projection of the left-hand side cartesian product and $$p'_{i}: prod_{j in I} A_{sigma(j)} to A_{sigma(i)}$$ the analogous projection for the right-hand side cartesian product (the $i$-projection assigns to every element $x$ in the respective cartesian product its $i$-component). Furthermore, let us define maps$$q_{i}=p_{sigma(i)},\ q'_{i}=p'_{sigma^{-1}(i)}$$

By the universality property of cartesian products, there will exist two maps
$$f: prod_{i in I} A_{i} to prod_{i in I} A_{sigma(i)},\ g: prod_{i in I} A_{sigma(i)} to prod_{i in I} A_{i}$$ unique with the respective properties that for every $i in I$ one has $p'_{i} circ f=q_i, p_i circ g=q'_{i}$. From this point on it is easy to verify that:
$$p_i circ g circ f=q'_i circ f=p'_{sigma^{-1}(i)} circ f=q_{sigma^{-1}(i)}=p_i$$ and similarly $$p'_i circ f circ g=q_i circ g=p_{sigma(i)} circ g=q'_{sigma(i)}=p'_i$$ for every index $i$, fact which entails the conclusion that $$f circ g=mathbf{1}_{prod_{i in I} A_{sigma(i)}},\ g circ f=mathbf{1}_{prod_{i in I} A_i}$$

In your particular case, this general result can be applied for a family of sets $C$ indexed by the set ${1, 2}$, such that $C_1=A, C_2=B$ together with the transposition $(12)$ (the only nontrivial permutation of the given two-element set, the one that swaps $1$ with $2$). According to the general argument, you obtain bijection $f: A times B to B times A$ given by the correspondence law $f(x,y)=(y,x)$. It is immediately verified that $f$ is its own inverse (what we call an involution), and that is the most direct way to settle the matter, rather than argue individually about injectivity and surjectivity.

share|cite|improve this answer

edited Mar 13 at 16:30

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

$endgroup$

share|cite|improve this answer

edited Mar 13 at 16:30

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436

answered Mar 13 at 7:36

ΑΘΩΑΘΩ

3436