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Proving that the union of two infinite disjoint sets with same cardinality is equipotent with either one
Clash Royale CLAN TAG#URR8PPP
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This question is almost a duplicate to the question Q, but I would like to prove it in a more personalised manner.
The fact, that the individual sets, say $A$ and $B$ are countable, as well as infinite, ensures the existence of some bijective function $f: mathbbN rightarrow A$ and $g: mathbbN rightarrow B$ .
Again, $A$ and $B$ being equipotent, there is some bijective function $h: A rightarrow B$.
Now, consider their union (keeping in mind that they are disjoint). Define a map $phi: mathbbN rightarrow A cup B $ such that,
$phi(n)= a_n/2 $ when $n$ is even; $b_n+1/ 2$ when $n$ is odd [ $A=a_n$ and $B=b_n$ (enumerability permits this) ] Evidently, this mapping is a bijection.
Now, we consider the composition $f^-1 phi: A rightarrow A cup B$ and $g^-1 phi: B rightarrow A cup B$. Both of them are bijective. Furthermore, the equipotency between the sets $A$ and $B$ implies that their union is equipotent to either one of them.
We now use this to prove that $mathbbQ$ and $mathbbQ^+$ are of same cardinality.
We now split $mathbbQ^*$ into $mathbbQ^+$ and $mathbbQ^-$ . Their elements are disjoint and both of them are equipotent. We use the previously proved proposition to deduce that $mathbbQ^*$ is equipotent to $mathbbQ^+$.
Is the above proof correct?
How do I extend the deduction to $mathbbQ$ ?
elementary-set-theory rational-numbers
edited Aug 2 at 12:59
zzuussee
1,152419
asked Aug 2 at 9:44
Subhasis Biswas
1749
add a comment |Â
up vote
0
down vote
favorite
This question is almost a duplicate to the question Q, but I would like to prove it in a more personalised manner.
The fact, that the individual sets, say $A$ and $B$ are countable, as well as infinite, ensures the existence of some bijective function $f: mathbbN rightarrow A$ and $g: mathbbN rightarrow B$ .
Again, $A$ and $B$ being equipotent, there is some bijective function $h: A rightarrow B$.
Now, consider their union (keeping in mind that they are disjoint). Define a map $phi: mathbbN rightarrow A cup B $ such that,
$phi(n)= a_n/2 $ when $n$ is even; $b_n+1/ 2$ when $n$ is odd [ $A=a_n$ and $B=b_n$ (enumerability permits this) ] Evidently, this mapping is a bijection.
Now, we consider the composition $f^-1 phi: A rightarrow A cup B$ and $g^-1 phi: B rightarrow A cup B$. Both of them are bijective. Furthermore, the equipotency between the sets $A$ and $B$ implies that their union is equipotent to either one of them.
We now use this to prove that $mathbbQ$ and $mathbbQ^+$ are of same cardinality.
We now split $mathbbQ^*$ into $mathbbQ^+$ and $mathbbQ^-$ . Their elements are disjoint and both of them are equipotent. We use the previously proved proposition to deduce that $mathbbQ^*$ is equipotent to $mathbbQ^+$.
Is the above proof correct?
How do I extend the deduction to $mathbbQ$ ?
elementary-set-theory rational-numbers
edited Aug 2 at 12:59
zzuussee
1,152419
asked Aug 2 at 9:44
Subhasis Biswas
1749
Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is almost a duplicate to the question Q, but I would like to prove it in a more personalised manner.
The fact, that the individual sets, say $A$ and $B$ are countable, as well as infinite, ensures the existence of some bijective function $f: mathbbN rightarrow A$ and $g: mathbbN rightarrow B$ .
Again, $A$ and $B$ being equipotent, there is some bijective function $h: A rightarrow B$.
Now, consider their union (keeping in mind that they are disjoint). Define a map $phi: mathbbN rightarrow A cup B $ such that,
$phi(n)= a_n/2 $ when $n$ is even; $b_n+1/ 2$ when $n$ is odd [ $A=a_n$ and $B=b_n$ (enumerability permits this) ] Evidently, this mapping is a bijection.
Now, we consider the composition $f^-1 phi: A rightarrow A cup B$ and $g^-1 phi: B rightarrow A cup B$. Both of them are bijective. Furthermore, the equipotency between the sets $A$ and $B$ implies that their union is equipotent to either one of them.
We now use this to prove that $mathbbQ$ and $mathbbQ^+$ are of same cardinality.
We now split $mathbbQ^*$ into $mathbbQ^+$ and $mathbbQ^-$ . Their elements are disjoint and both of them are equipotent. We use the previously proved proposition to deduce that $mathbbQ^*$ is equipotent to $mathbbQ^+$.
Is the above proof correct?
How do I extend the deduction to $mathbbQ$ ?
elementary-set-theory rational-numbers
edited Aug 2 at 12:59
zzuussee
1,152419
asked Aug 2 at 9:44
Subhasis Biswas
1749
This question is almost a duplicate to the question Q, but I would like to prove it in a more personalised manner.
The fact, that the individual sets, say $A$ and $B$ are countable, as well as infinite, ensures the existence of some bijective function $f: mathbbN rightarrow A$ and $g: mathbbN rightarrow B$ .
Again, $A$ and $B$ being equipotent, there is some bijective function $h: A rightarrow B$.
Now, consider their union (keeping in mind that they are disjoint). Define a map $phi: mathbbN rightarrow A cup B $ such that,
$phi(n)= a_n/2 $ when $n$ is even; $b_n+1/ 2$ when $n$ is odd [ $A=a_n$ and $B=b_n$ (enumerability permits this) ] Evidently, this mapping is a bijection.
Now, we consider the composition $f^-1 phi: A rightarrow A cup B$ and $g^-1 phi: B rightarrow A cup B$. Both of them are bijective. Furthermore, the equipotency between the sets $A$ and $B$ implies that their union is equipotent to either one of them.
We now use this to prove that $mathbbQ$ and $mathbbQ^+$ are of same cardinality.
We now split $mathbbQ^*$ into $mathbbQ^+$ and $mathbbQ^-$ . Their elements are disjoint and both of them are equipotent. We use the previously proved proposition to deduce that $mathbbQ^*$ is equipotent to $mathbbQ^+$.
Is the above proof correct?
How do I extend the deduction to $mathbbQ$ ?
elementary-set-theory rational-numbers
edited Aug 2 at 12:59
zzuussee
1,152419
asked Aug 2 at 9:44
Subhasis Biswas
1749
edited Aug 2 at 12:59
zzuussee
1,152419
edited Aug 2 at 12:59
zzuussee
1,152419
edited Aug 2 at 12:59
zzuussee
1,152419
1,152419
asked Aug 2 at 9:44
Subhasis Biswas
1749
asked Aug 2 at 9:44
Subhasis Biswas
1749
asked Aug 2 at 9:44
Subhasis Biswas
1749
1749
Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56
add a comment |Â
Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56
Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56
add a comment |Â
1 Answer
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up vote
1
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I first want to note, that in your first comment where you suggest a bijection from $0/1,0/2,0/3,dots$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $0/1,0/2,0/3,dots=0$ which is most certainly not bijective with $mathbbN$.
The proof that for countable disjoint $A,B$, there is a bijection between $Acup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.
This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $mathbbQ$ and $mathbbQ^+$:
- You may show that $mathbbQ$ and $mathbbQ^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:mathbbNtomathbbQ$ and $g:mathbbNtomathbbQ^+$. Then $h:mathbbQtomathbbQ^+$, $qto g(f^-1(q))$ may be checked by you to be bijective.
- You can proceed by splitting $mathbbQ=mathbbQ^-cup(mathbbQ^+cup0)$ or $mathbbQ=(mathbbQ^-cup0)cupmathbbQ^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.
As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.
answered Aug 2 at 13:06
zzuussee
1,152419
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I first want to note, that in your first comment where you suggest a bijection from $0/1,0/2,0/3,dots$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $0/1,0/2,0/3,dots=0$ which is most certainly not bijective with $mathbbN$.
The proof that for countable disjoint $A,B$, there is a bijection between $Acup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.
This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $mathbbQ$ and $mathbbQ^+$:
- You may show that $mathbbQ$ and $mathbbQ^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:mathbbNtomathbbQ$ and $g:mathbbNtomathbbQ^+$. Then $h:mathbbQtomathbbQ^+$, $qto g(f^-1(q))$ may be checked by you to be bijective.
- You can proceed by splitting $mathbbQ=mathbbQ^-cup(mathbbQ^+cup0)$ or $mathbbQ=(mathbbQ^-cup0)cupmathbbQ^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.
As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.
answered Aug 2 at 13:06
zzuussee
1,152419
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
add a comment |Â
up vote
1
down vote
accepted
I first want to note, that in your first comment where you suggest a bijection from $0/1,0/2,0/3,dots$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $0/1,0/2,0/3,dots=0$ which is most certainly not bijective with $mathbbN$.
The proof that for countable disjoint $A,B$, there is a bijection between $Acup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.
This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $mathbbQ$ and $mathbbQ^+$:
- You may show that $mathbbQ$ and $mathbbQ^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:mathbbNtomathbbQ$ and $g:mathbbNtomathbbQ^+$. Then $h:mathbbQtomathbbQ^+$, $qto g(f^-1(q))$ may be checked by you to be bijective.
- You can proceed by splitting $mathbbQ=mathbbQ^-cup(mathbbQ^+cup0)$ or $mathbbQ=(mathbbQ^-cup0)cupmathbbQ^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.
As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.
answered Aug 2 at 13:06
zzuussee
1,152419
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I first want to note, that in your first comment where you suggest a bijection from $0/1,0/2,0/3,dots$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $0/1,0/2,0/3,dots=0$ which is most certainly not bijective with $mathbbN$.
The proof that for countable disjoint $A,B$, there is a bijection between $Acup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.
This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $mathbbQ$ and $mathbbQ^+$:
- You may show that $mathbbQ$ and $mathbbQ^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:mathbbNtomathbbQ$ and $g:mathbbNtomathbbQ^+$. Then $h:mathbbQtomathbbQ^+$, $qto g(f^-1(q))$ may be checked by you to be bijective.
- You can proceed by splitting $mathbbQ=mathbbQ^-cup(mathbbQ^+cup0)$ or $mathbbQ=(mathbbQ^-cup0)cupmathbbQ^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.
As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.
answered Aug 2 at 13:06
zzuussee
1,152419
I first want to note, that in your first comment where you suggest a bijection from $0/1,0/2,0/3,dots$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $0/1,0/2,0/3,dots=0$ which is most certainly not bijective with $mathbbN$.
The proof that for countable disjoint $A,B$, there is a bijection between $Acup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.
This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $mathbbQ$ and $mathbbQ^+$:
- You may show that $mathbbQ$ and $mathbbQ^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:mathbbNtomathbbQ$ and $g:mathbbNtomathbbQ^+$. Then $h:mathbbQtomathbbQ^+$, $qto g(f^-1(q))$ may be checked by you to be bijective.
- You can proceed by splitting $mathbbQ=mathbbQ^-cup(mathbbQ^+cup0)$ or $mathbbQ=(mathbbQ^-cup0)cupmathbbQ^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.
As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.
answered Aug 2 at 13:06
zzuussee
1,152419
edited Aug 2 at 13:22
answered Aug 2 at 13:06
zzuussee
1,152419
answered Aug 2 at 13:06
zzuussee
1,152419
answered Aug 2 at 13:06
zzuussee
1,152419
1,152419
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
add a comment |Â
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
Finally someone read it. Such a nice and insightful answer!!!
– Subhasis Biswas
Aug 2 at 14:19
add a comment |Â
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Well, we can define an unconventional bijection from $0/1, 0/2, 0/3....$ to $mathbbN$, but this doesn't seem to be valid.
– Subhasis Biswas
Aug 2 at 10:03
Other workaround might be splitting the set $mathbbQ$ into $mathbbQ^-$ and $mathbbQ^+ cup 0$. Setting $b_1=0$ in the second set solves the problem, I guess.
– Subhasis Biswas
Aug 2 at 10:56