Equal Cardinality and Bijection Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Set Related Questiontroubles proving every subset of a finite set is finite with naive set theoryConfusion about cofinalityAny two sets $Y$, $Z$ have the same cardinality $iff$ there are injective functions $f: Y rightarrow Z$ and $g: Z rightarrow Y$.If $X$ is a finite set and $Y subset X$. Prove that $Y$ is finite.Prove that $f: A to B$ is surjective iff $forall b in B, ((A times $$b$$) cap G_f) neq emptyset $Basic subset question, confused“one-to-one correspondence”(bijection) and the size of two infinite setsExamples of bijections from $mathbb Ztomathbb Z$ such that their sum is a bijection?Am I taking the “true meaning” of “$subset$” and in the context of this question to be wrong? Or the question is incorrect itself?
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Equal Cardinality and Bijection
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Set Related Questiontroubles proving every subset of a finite set is finite with naive set theoryConfusion about cofinalityAny two sets $Y$, $Z$ have the same cardinality $iff$ there are injective functions $f: Y rightarrow Z$ and $g: Z rightarrow Y$.If $X$ is a finite set and $Y subset X$. Prove that $Y$ is finite.Prove that $f: A to B$ is surjective iff $forall b in B, ((A times $$b$$) cap G_f) neq emptyset $Basic subset question, confused“one-to-one correspondence”(bijection) and the size of two infinite setsExamples of bijections from $mathbb Ztomathbb Z$ such that their sum is a bijection?Am I taking the “true meaning” of “$subset$” and in the context of this question to be wrong? Or the question is incorrect itself?
$begingroup$
I ran into the following question:
Suppose that $B subset A$ and that $exists$ a bijection $f: Amapsto B$. What may be reasonably deduced about $A$?
I think either there is something wrong with the question because the existence of a bijection is the formal condition for equal cardinality of two sets. But $mathrm cardA=mathrm cardB$ and $Bsubset A$ being simultaneous conditions is rather absurd to me. Maybe $emptyset$ has something to do here.
Is there another perspective that helps in understanding this better? Also, what am I missing? Thanks
functions elementary-set-theory
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
$endgroup$
add a comment |
$begingroup$
I ran into the following question:
Suppose that $B subset A$ and that $exists$ a bijection $f: Amapsto B$. What may be reasonably deduced about $A$?
I think either there is something wrong with the question because the existence of a bijection is the formal condition for equal cardinality of two sets. But $mathrm cardA=mathrm cardB$ and $Bsubset A$ being simultaneous conditions is rather absurd to me. Maybe $emptyset$ has something to do here.
Is there another perspective that helps in understanding this better? Also, what am I missing? Thanks
functions elementary-set-theory
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
$endgroup$
2
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
1
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48
add a comment |
$begingroup$
I ran into the following question:
Suppose that $B subset A$ and that $exists$ a bijection $f: Amapsto B$. What may be reasonably deduced about $A$?
I think either there is something wrong with the question because the existence of a bijection is the formal condition for equal cardinality of two sets. But $mathrm cardA=mathrm cardB$ and $Bsubset A$ being simultaneous conditions is rather absurd to me. Maybe $emptyset$ has something to do here.
Is there another perspective that helps in understanding this better? Also, what am I missing? Thanks
functions elementary-set-theory
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
$endgroup$
I ran into the following question:
Suppose that $B subset A$ and that $exists$ a bijection $f: Amapsto B$. What may be reasonably deduced about $A$?
I think either there is something wrong with the question because the existence of a bijection is the formal condition for equal cardinality of two sets. But $mathrm cardA=mathrm cardB$ and $Bsubset A$ being simultaneous conditions is rather absurd to me. Maybe $emptyset$ has something to do here.
Is there another perspective that helps in understanding this better? Also, what am I missing? Thanks
functions elementary-set-theory
functions elementary-set-theory
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
asked Apr 2 at 17:41
Paras KhoslaParas Khosla
3,3121627
3,3121627
2
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
1
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48
add a comment |
2
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
1
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48
2
2
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
1
1
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This condition is the definition of an infinite set. $A$ is infinite iff $exists$ a bijection between $A$ and its proper subset, that in this case is $B$. So it can be concluded that $A$ is an infinite set.
As an example, consider a bijection $f$ from $A=x: xin mathbbZ$ and $B=x: x=2n, ninmathbbZ$. Using simple notation, $f:Amapsto B mid f(n)=2n$.
$endgroup$
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
add a comment |
Your Answer
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$begingroup$
This condition is the definition of an infinite set. $A$ is infinite iff $exists$ a bijection between $A$ and its proper subset, that in this case is $B$. So it can be concluded that $A$ is an infinite set.
As an example, consider a bijection $f$ from $A=x: xin mathbbZ$ and $B=x: x=2n, ninmathbbZ$. Using simple notation, $f:Amapsto B mid f(n)=2n$.
$endgroup$
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
add a comment |
$begingroup$
This condition is the definition of an infinite set. $A$ is infinite iff $exists$ a bijection between $A$ and its proper subset, that in this case is $B$. So it can be concluded that $A$ is an infinite set.
As an example, consider a bijection $f$ from $A=x: xin mathbbZ$ and $B=x: x=2n, ninmathbbZ$. Using simple notation, $f:Amapsto B mid f(n)=2n$.
$endgroup$
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
add a comment |
$begingroup$
This condition is the definition of an infinite set. $A$ is infinite iff $exists$ a bijection between $A$ and its proper subset, that in this case is $B$. So it can be concluded that $A$ is an infinite set.
As an example, consider a bijection $f$ from $A=x: xin mathbbZ$ and $B=x: x=2n, ninmathbbZ$. Using simple notation, $f:Amapsto B mid f(n)=2n$.
$endgroup$
This condition is the definition of an infinite set. $A$ is infinite iff $exists$ a bijection between $A$ and its proper subset, that in this case is $B$. So it can be concluded that $A$ is an infinite set.
As an example, consider a bijection $f$ from $A=x: xin mathbbZ$ and $B=x: x=2n, ninmathbbZ$. Using simple notation, $f:Amapsto B mid f(n)=2n$.
answered Apr 2 at 17:58
community wiki
Paras Khosla
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
add a comment |
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
1
1
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets.
$endgroup$
– Cameron Buie
Apr 2 at 18:27
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
$begingroup$
@CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :)
$endgroup$
– Paras Khosla
Apr 2 at 18:43
1
1
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact.
$endgroup$
– Cameron Buie
Apr 2 at 18:51
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
$begingroup$
When you put it like that, it certainly is insightful. Thanks :))
$endgroup$
– Paras Khosla
Apr 2 at 18:53
add a comment |
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2
$begingroup$
If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$?
$endgroup$
– Thomas Andrews
Apr 2 at 17:43
$begingroup$
I get it because the $mathrmcardmathrm Z$ and $mathrmcard B$ where $B=x: x=2n, nin mathrm Z$ are both equal to $infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set?
$endgroup$
– Paras Khosla
Apr 2 at 17:46
1
$begingroup$
You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset.
$endgroup$
– SMM
Apr 2 at 17:47
$begingroup$
Thanks for clarifying @SMM
$endgroup$
– Paras Khosla
Apr 2 at 17:48