About a bijective function with the disjoint sets
$begingroup$
In this video by 0:55, it says
Lemma: Let $A,B,C,D$ be sets with $Acap B=emptyset$ and $Ccap D=emptyset$. Suppose that $F_1:Ato C$ and $F_2:Bto D$ are both bijections. Define $F:Acup Bto Ccup D$ by
$$
F(x)=begin{cases}
F_1(x) & text{ if } xin A \
F_2(x) & text{ if } xin B
end{cases}
$$
Then $F$ is a bijection.
To me, it is easy to prove it by looking at the $F|_A$ and $F|_B$ seperately, and then, by construction, $F$ is bijective by collecting the elements of $A$ and $B$. The question is, why do the intersections $Acap B$ and $Ccap D$ have to be empty, in particular the last one?
discrete-mathematics
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
$endgroup$
add a comment |
$begingroup$
In this video by 0:55, it says
Lemma: Let $A,B,C,D$ be sets with $Acap B=emptyset$ and $Ccap D=emptyset$. Suppose that $F_1:Ato C$ and $F_2:Bto D$ are both bijections. Define $F:Acup Bto Ccup D$ by
$$
F(x)=begin{cases}
F_1(x) & text{ if } xin A \
F_2(x) & text{ if } xin B
end{cases}
$$
Then $F$ is a bijection.
To me, it is easy to prove it by looking at the $F|_A$ and $F|_B$ seperately, and then, by construction, $F$ is bijective by collecting the elements of $A$ and $B$. The question is, why do the intersections $Acap B$ and $Ccap D$ have to be empty, in particular the last one?
discrete-mathematics
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
$endgroup$
add a comment |
$begingroup$
In this video by 0:55, it says
Lemma: Let $A,B,C,D$ be sets with $Acap B=emptyset$ and $Ccap D=emptyset$. Suppose that $F_1:Ato C$ and $F_2:Bto D$ are both bijections. Define $F:Acup Bto Ccup D$ by
$$
F(x)=begin{cases}
F_1(x) & text{ if } xin A \
F_2(x) & text{ if } xin B
end{cases}
$$
Then $F$ is a bijection.
To me, it is easy to prove it by looking at the $F|_A$ and $F|_B$ seperately, and then, by construction, $F$ is bijective by collecting the elements of $A$ and $B$. The question is, why do the intersections $Acap B$ and $Ccap D$ have to be empty, in particular the last one?
discrete-mathematics
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
$endgroup$
In this video by 0:55, it says
Lemma: Let $A,B,C,D$ be sets with $Acap B=emptyset$ and $Ccap D=emptyset$. Suppose that $F_1:Ato C$ and $F_2:Bto D$ are both bijections. Define $F:Acup Bto Ccup D$ by
$$
F(x)=begin{cases}
F_1(x) & text{ if } xin A \
F_2(x) & text{ if } xin B
end{cases}
$$
Then $F$ is a bijection.
To me, it is easy to prove it by looking at the $F|_A$ and $F|_B$ seperately, and then, by construction, $F$ is bijective by collecting the elements of $A$ and $B$. The question is, why do the intersections $Acap B$ and $Ccap D$ have to be empty, in particular the last one?
discrete-mathematics
discrete-mathematics
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
asked Dec 2 '18 at 3:24
UnknownWUnknownW
1,001922
1,001922
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the intersection $A cap B$ is nonempty, the function may not be well-defined — if $xin A cap B$ and $F_1(x) neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C cap D$ is nonempty, the function may not be an injection. For $y in C cap D$, there must be $x_1 in A$ and $x_2 in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 neq x_2$ (which may very well be true), then $F$ is not an injection.
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
$endgroup$
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
|
show 3 more comments
$begingroup$
If $y in Acap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $Ccap D = emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)
Likewise if $z in Ccap D$ then it's possible there is an $a in A$ and a $b in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
votes
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$begingroup$
If the intersection $A cap B$ is nonempty, the function may not be well-defined — if $xin A cap B$ and $F_1(x) neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C cap D$ is nonempty, the function may not be an injection. For $y in C cap D$, there must be $x_1 in A$ and $x_2 in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 neq x_2$ (which may very well be true), then $F$ is not an injection.
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
$endgroup$
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
|
show 3 more comments
$begingroup$
If the intersection $A cap B$ is nonempty, the function may not be well-defined — if $xin A cap B$ and $F_1(x) neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C cap D$ is nonempty, the function may not be an injection. For $y in C cap D$, there must be $x_1 in A$ and $x_2 in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 neq x_2$ (which may very well be true), then $F$ is not an injection.
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
$endgroup$
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
|
show 3 more comments
$begingroup$
If the intersection $A cap B$ is nonempty, the function may not be well-defined — if $xin A cap B$ and $F_1(x) neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C cap D$ is nonempty, the function may not be an injection. For $y in C cap D$, there must be $x_1 in A$ and $x_2 in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 neq x_2$ (which may very well be true), then $F$ is not an injection.
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
$endgroup$
If the intersection $A cap B$ is nonempty, the function may not be well-defined — if $xin A cap B$ and $F_1(x) neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C cap D$ is nonempty, the function may not be an injection. For $y in C cap D$, there must be $x_1 in A$ and $x_2 in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 neq x_2$ (which may very well be true), then $F$ is not an injection.
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
edited Dec 2 '18 at 4:22
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
answered Dec 2 '18 at 3:28
plattyplatty
3,370320
3,370320
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
|
show 3 more comments
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
$begingroup$
I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:33
1
1
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
@BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x in A cap B$ such that $F_1(x) neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold.
$endgroup$
– platty
Dec 2 '18 at 3:37
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
$begingroup$
I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined.
$endgroup$
– Bertrand Wittgenstein's Ghost
Dec 2 '18 at 3:42
1
1
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
$begingroup$
Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand?
$endgroup$
– platty
Dec 2 '18 at 3:45
1
1
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
$begingroup$
By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) neq F_2(x)$, the function $F$ is not well-defined.
$endgroup$
– platty
Dec 2 '18 at 4:09
|
show 3 more comments
$begingroup$
If $y in Acap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $Ccap D = emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)
Likewise if $z in Ccap D$ then it's possible there is an $a in A$ and a $b in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
$endgroup$
add a comment |
$begingroup$
If $y in Acap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $Ccap D = emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)
Likewise if $z in Ccap D$ then it's possible there is an $a in A$ and a $b in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
$endgroup$
add a comment |
$begingroup$
If $y in Acap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $Ccap D = emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)
Likewise if $z in Ccap D$ then it's possible there is an $a in A$ and a $b in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
$endgroup$
If $y in Acap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $Ccap D = emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)
Likewise if $z in Ccap D$ then it's possible there is an $a in A$ and a $b in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
answered Dec 2 '18 at 4:53
fleabloodfleablood
71k22686
71k22686
add a comment |
add a comment |
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