Set theory proof explanation (union of sets) [closed]
$begingroup$
I am practicing for my exam tomorrow and I stumbled upon an exercise I cannot solve
Prove that:
$cup A_i times cup B_j = cup(A_i times B_j)$
$i in I$, $j in J$,$(I,j) in Itimes J$
Can someone help me with that?
elementary-set-theory
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
$endgroup$
closed as off-topic by José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo Dec 18 '18 at 10:35
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." – José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am practicing for my exam tomorrow and I stumbled upon an exercise I cannot solve
Prove that:
$cup A_i times cup B_j = cup(A_i times B_j)$
$i in I$, $j in J$,$(I,j) in Itimes J$
Can someone help me with that?
elementary-set-theory
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
$endgroup$
closed as off-topic by José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo Dec 18 '18 at 10:35
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." – José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14
add a comment |
$begingroup$
I am practicing for my exam tomorrow and I stumbled upon an exercise I cannot solve
Prove that:
$cup A_i times cup B_j = cup(A_i times B_j)$
$i in I$, $j in J$,$(I,j) in Itimes J$
Can someone help me with that?
elementary-set-theory
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
$endgroup$
I am practicing for my exam tomorrow and I stumbled upon an exercise I cannot solve
Prove that:
$cup A_i times cup B_j = cup(A_i times B_j)$
$i in I$, $j in J$,$(I,j) in Itimes J$
Can someone help me with that?
elementary-set-theory
elementary-set-theory
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
edited Dec 18 '18 at 0:50
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
asked Dec 18 '18 at 0:09
mrnobodymrnobody
14818
14818
closed as off-topic by José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo Dec 18 '18 at 10:35
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." – José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo Dec 18 '18 at 10:35
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." – José Carlos Santos, Eevee Trainer, Saad, Carl Schildkraut, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14
add a comment |
1
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14
1
1
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Glossing over trivialities by assuming $left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$ is non-empty, here is one direction of the double containment argument:
Suppose $(x,y) in left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$. By the definition of cartesian product we have $x in bigcup_{i in I} A_i$ and $y in bigcup_{j in J} B_j$. By the definition of set union we know there is $k in I$ so that $x in A_k$. Likewise there is $l in J$ so that $y in B_l$. And so we have $(x,y) in A_k times B_l$ with $(k,l) in I times J$. Therefore $(x,y) in bigcup_{(i,j) in I times J} left( A_i times B_j right)$.
It might be helpful to notice that the two hyperlinked definitions are fairly important here.
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Glossing over trivialities by assuming $left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$ is non-empty, here is one direction of the double containment argument:
Suppose $(x,y) in left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$. By the definition of cartesian product we have $x in bigcup_{i in I} A_i$ and $y in bigcup_{j in J} B_j$. By the definition of set union we know there is $k in I$ so that $x in A_k$. Likewise there is $l in J$ so that $y in B_l$. And so we have $(x,y) in A_k times B_l$ with $(k,l) in I times J$. Therefore $(x,y) in bigcup_{(i,j) in I times J} left( A_i times B_j right)$.
It might be helpful to notice that the two hyperlinked definitions are fairly important here.
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
add a comment |
$begingroup$
Glossing over trivialities by assuming $left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$ is non-empty, here is one direction of the double containment argument:
Suppose $(x,y) in left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$. By the definition of cartesian product we have $x in bigcup_{i in I} A_i$ and $y in bigcup_{j in J} B_j$. By the definition of set union we know there is $k in I$ so that $x in A_k$. Likewise there is $l in J$ so that $y in B_l$. And so we have $(x,y) in A_k times B_l$ with $(k,l) in I times J$. Therefore $(x,y) in bigcup_{(i,j) in I times J} left( A_i times B_j right)$.
It might be helpful to notice that the two hyperlinked definitions are fairly important here.
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
add a comment |
$begingroup$
Glossing over trivialities by assuming $left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$ is non-empty, here is one direction of the double containment argument:
Suppose $(x,y) in left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$. By the definition of cartesian product we have $x in bigcup_{i in I} A_i$ and $y in bigcup_{j in J} B_j$. By the definition of set union we know there is $k in I$ so that $x in A_k$. Likewise there is $l in J$ so that $y in B_l$. And so we have $(x,y) in A_k times B_l$ with $(k,l) in I times J$. Therefore $(x,y) in bigcup_{(i,j) in I times J} left( A_i times B_j right)$.
It might be helpful to notice that the two hyperlinked definitions are fairly important here.
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
Glossing over trivialities by assuming $left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$ is non-empty, here is one direction of the double containment argument:
Suppose $(x,y) in left( bigcup_{i in I} A_i right) times left( bigcup_{j in J} B_j right)$. By the definition of cartesian product we have $x in bigcup_{i in I} A_i$ and $y in bigcup_{j in J} B_j$. By the definition of set union we know there is $k in I$ so that $x in A_k$. Likewise there is $l in J$ so that $y in B_l$. And so we have $(x,y) in A_k times B_l$ with $(k,l) in I times J$. Therefore $(x,y) in bigcup_{(i,j) in I times J} left( A_i times B_j right)$.
It might be helpful to notice that the two hyperlinked definitions are fairly important here.
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
answered Dec 18 '18 at 1:02
Matt A PeltoMatt A Pelto
2,622621
2,622621
add a comment |
add a comment |
1
$begingroup$
Show that any element of the LHS is in the RHS, and conversely. Post what you have done for a start and explain where you are stuck, then someone may be able to help you get a bit further. You will need to correct the typos in the question first.
$endgroup$
– David
Dec 18 '18 at 0:14