Mixing
Find all injections from set ${1,2,3}$ into set ${4,5}$
-4
$begingroup$
I am supposed to find all the injections from set ${1,2,3}$ to set ${4,5}$. Is it possible to do (meaning are there any injections?)? If so, what are the possible injections?
functions elementary-set-theory
edited Feb 19 at 0:14
Heptapod
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
$endgroup$
add a comment |
-4
$begingroup$
I am supposed to find all the injections from set ${1,2,3}$ to set ${4,5}$. Is it possible to do (meaning are there any injections?)? If so, what are the possible injections?
functions elementary-set-theory
edited Feb 19 at 0:14
Heptapod
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
$endgroup$
1
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35
add a comment |
-4
-4
-4
$begingroup$
I am supposed to find all the injections from set ${1,2,3}$ to set ${4,5}$. Is it possible to do (meaning are there any injections?)? If so, what are the possible injections?
functions elementary-set-theory
edited Feb 19 at 0:14
Heptapod
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
$endgroup$
I am supposed to find all the injections from set ${1,2,3}$ to set ${4,5}$. Is it possible to do (meaning are there any injections?)? If so, what are the possible injections?
functions elementary-set-theory
functions elementary-set-theory
edited Feb 19 at 0:14
Heptapod
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
edited Feb 19 at 0:14
Heptapod
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
edited Feb 19 at 0:14
Heptapod
716317
edited Feb 19 at 0:14
Heptapod
716317
edited Feb 19 at 0:14
Heptapod
716317
716317
asked Jan 25 at 17:32
J. DoeJ. Doe
61
asked Jan 25 at 17:32
J. DoeJ. Doe
61
asked Jan 25 at 17:32
J. DoeJ. Doe
61
61
1
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35
add a comment |
1
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35
1
1
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
Hint:
If $f:Alongrightarrow B$ is an injection, then $:|A|le |B|$.
answered Jan 25 at 17:37
BernardBernard
123k741116
$endgroup$
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
add a comment |
0
$begingroup$
It is not possible.
An injection is a function where each output is mapped to by at most one input. Since it is a function, every input must map to one output. However, you have more inputs than outputs, so it is not possible for each input to map to different outputs (see Pidgeonhole Principle).
answered Jan 25 at 17:35
JupiterianJupiterian
93
$endgroup$
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint:
If $f:Alongrightarrow B$ is an injection, then $:|A|le |B|$.
answered Jan 25 at 17:37
BernardBernard
123k741116
$endgroup$
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
add a comment |
1
$begingroup$
Hint:
If $f:Alongrightarrow B$ is an injection, then $:|A|le |B|$.
answered Jan 25 at 17:37
BernardBernard
123k741116
$endgroup$
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
add a comment |
1
1
1
$begingroup$
Hint:
If $f:Alongrightarrow B$ is an injection, then $:|A|le |B|$.
answered Jan 25 at 17:37
BernardBernard
123k741116
$endgroup$
Hint:
If $f:Alongrightarrow B$ is an injection, then $:|A|le |B|$.
answered Jan 25 at 17:37
BernardBernard
123k741116
answered Jan 25 at 17:37
BernardBernard
123k741116
answered Jan 25 at 17:37
BernardBernard
123k741116
answered Jan 25 at 17:37
BernardBernard
123k741116
123k741116
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
add a comment |
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
Thanks. And what about the bijection? Is it possible?
$endgroup$
– J. Doe
Jan 25 at 17:39
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
bijection $=$ injection $+$ surjection.
$endgroup$
– Bernard
Jan 25 at 17:41
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
$begingroup$
@J.Doe a bijection is a special kind of injection (one which is also a surjection). If there are no injections, then certainly there are no bijections either. For finite sets, there exists at least one bijection between two sets if and only if they have the same number of elements. This extends to infinite sets as well but we have to be careful about what we mean by an infinite set's "number of elements." Your sets have a different number of elements (three and two respectively) and so clearly have no bijections between them.
$endgroup$
– JMoravitz
Jan 25 at 17:42
add a comment |
0
$begingroup$
It is not possible.
An injection is a function where each output is mapped to by at most one input. Since it is a function, every input must map to one output. However, you have more inputs than outputs, so it is not possible for each input to map to different outputs (see Pidgeonhole Principle).
answered Jan 25 at 17:35
JupiterianJupiterian
93
$endgroup$
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
add a comment |
0
$begingroup$
It is not possible.
An injection is a function where each output is mapped to by at most one input. Since it is a function, every input must map to one output. However, you have more inputs than outputs, so it is not possible for each input to map to different outputs (see Pidgeonhole Principle).
answered Jan 25 at 17:35
JupiterianJupiterian
93
$endgroup$
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
add a comment |
0
0
0
$begingroup$
It is not possible.
An injection is a function where each output is mapped to by at most one input. Since it is a function, every input must map to one output. However, you have more inputs than outputs, so it is not possible for each input to map to different outputs (see Pidgeonhole Principle).
answered Jan 25 at 17:35
JupiterianJupiterian
93
$endgroup$
It is not possible.
An injection is a function where each output is mapped to by at most one input. Since it is a function, every input must map to one output. However, you have more inputs than outputs, so it is not possible for each input to map to different outputs (see Pidgeonhole Principle).
answered Jan 25 at 17:35
JupiterianJupiterian
93
edited Jan 25 at 17:38
answered Jan 25 at 17:35
JupiterianJupiterian
93
answered Jan 25 at 17:35
JupiterianJupiterian
93
answered Jan 25 at 17:35
JupiterianJupiterian
93
93
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
add a comment |
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
Thanks! and what about the bijection of the same sets?
$endgroup$
– J. Doe
Jan 25 at 17:36
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
A function is such that for each input, the function outputs at most one output. An injection is such that, for $x, y in$ the domain of the function, if $f(x) = f(y)$ then $x = y$.
$endgroup$
– Namaste
Jan 25 at 17:38
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
$begingroup$
Thanks, I added that to the answer, and clarified a bit.
$endgroup$
– Jupiterian
Jan 25 at 17:39
add a comment |
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1
$begingroup$
Please include in your post what you understand to be the definition of an injection from one set to another.
$endgroup$
– Namaste
Jan 25 at 17:35