On injective, bijective, surjective
$begingroup$
Consider a function $f: mathcal{A}rightarrow mathbb{R}$ where $mathcal{A}equiv {a_1,a_2,a_3,a_4}subset mathbb{R}$ .
We know that $f$ is injective, i.e., $f(a_j)neq f(a_k)$ $forall a_jneq a_k$.
I would like your help with the following: given that the domain of $f$ is finite (and hence the image set of $f$ is finite), does this mean that $f$ being injective implies $f$ being bijective? If not, can you give an example of $f$ with domain $mathcal{A}$ that is injective but not bijective?
functional-analysis functions
edited Jan 11 at 20:26
user3482749
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
$endgroup$
add a comment |
$begingroup$
Consider a function $f: mathcal{A}rightarrow mathbb{R}$ where $mathcal{A}equiv {a_1,a_2,a_3,a_4}subset mathbb{R}$ .
We know that $f$ is injective, i.e., $f(a_j)neq f(a_k)$ $forall a_jneq a_k$.
I would like your help with the following: given that the domain of $f$ is finite (and hence the image set of $f$ is finite), does this mean that $f$ being injective implies $f$ being bijective? If not, can you give an example of $f$ with domain $mathcal{A}$ that is injective but not bijective?
functional-analysis functions
edited Jan 11 at 20:26
user3482749
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
$endgroup$
4
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
1
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30
add a comment |
$begingroup$
Consider a function $f: mathcal{A}rightarrow mathbb{R}$ where $mathcal{A}equiv {a_1,a_2,a_3,a_4}subset mathbb{R}$ .
We know that $f$ is injective, i.e., $f(a_j)neq f(a_k)$ $forall a_jneq a_k$.
I would like your help with the following: given that the domain of $f$ is finite (and hence the image set of $f$ is finite), does this mean that $f$ being injective implies $f$ being bijective? If not, can you give an example of $f$ with domain $mathcal{A}$ that is injective but not bijective?
functional-analysis functions
edited Jan 11 at 20:26
user3482749
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
$endgroup$
Consider a function $f: mathcal{A}rightarrow mathbb{R}$ where $mathcal{A}equiv {a_1,a_2,a_3,a_4}subset mathbb{R}$ .
We know that $f$ is injective, i.e., $f(a_j)neq f(a_k)$ $forall a_jneq a_k$.
I would like your help with the following: given that the domain of $f$ is finite (and hence the image set of $f$ is finite), does this mean that $f$ being injective implies $f$ being bijective? If not, can you give an example of $f$ with domain $mathcal{A}$ that is injective but not bijective?
functional-analysis functions
functional-analysis functions
edited Jan 11 at 20:26
user3482749
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
edited Jan 11 at 20:26
user3482749
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
edited Jan 11 at 20:26
user3482749
4,3291119
edited Jan 11 at 20:26
user3482749
4,3291119
edited Jan 11 at 20:26
user3482749
4,3291119
4,3291119
asked Jan 11 at 20:25
STFSTF
571422
asked Jan 11 at 20:25
STFSTF
571422
asked Jan 11 at 20:25
STFSTF
571422
571422
4
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
1
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30
add a comment |
4
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
1
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30
4
4
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
1
1
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: mathcal{A}tomathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $mathcal{A}tomathbb{R}$ would work equally.
However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:
$f$ is injective.
$f$ is surjective.
$f$ is bijective.
To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
$endgroup$
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
add a comment |
$begingroup$
The map $f$ is definitely not surjective because it misses everything in $mathbb{R}setminus{f(a_{1}),f(a_{2}),f(a_{3}),f(a_{4})}$
answered Jan 11 at 20:28
pwerthpwerth
3,340417
$endgroup$
add a comment |
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$begingroup$
Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: mathcal{A}tomathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $mathcal{A}tomathbb{R}$ would work equally.
However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:
$f$ is injective.
$f$ is surjective.
$f$ is bijective.
To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
$endgroup$
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
add a comment |
$begingroup$
Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: mathcal{A}tomathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $mathcal{A}tomathbb{R}$ would work equally.
However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:
$f$ is injective.
$f$ is surjective.
$f$ is bijective.
To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
$endgroup$
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
add a comment |
$begingroup$
Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: mathcal{A}tomathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $mathcal{A}tomathbb{R}$ would work equally.
However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:
$f$ is injective.
$f$ is surjective.
$f$ is bijective.
To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
$endgroup$
Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: mathcal{A}tomathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $mathcal{A}tomathbb{R}$ would work equally.
However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:
$f$ is injective.
$f$ is surjective.
$f$ is bijective.
To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
edited Jan 11 at 20:33
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
answered Jan 11 at 20:28
user3482749user3482749
4,3291119
4,3291119
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
add a comment |
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
Thanks. Then I think that I have a more basic question on the definition of codomain of a function. And I apologise in advance for my very untechnical language. While the image set of a function is uniquely define, is there a "rule of thumb" to define the codomain of a function? E.g. $f=x+2$ with domain the set of natural number $mathbb{N}$ could have as codomain $mathbb{R}$ or also $mathbb{N}$. In turn, the classification as injective, surjective depends on how we choose the codomain.
$endgroup$
– STF
Jan 11 at 20:33
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
$begingroup$
The codomain of a function is, like the domain, part of the definition of the function. $f = x + 2$ does not define a function, by itself: you also need to specify both the domain and the codomain. As you've noticed, changing either of these changes the function.
$endgroup$
– user3482749
Jan 11 at 20:35
add a comment |
$begingroup$
The map $f$ is definitely not surjective because it misses everything in $mathbb{R}setminus{f(a_{1}),f(a_{2}),f(a_{3}),f(a_{4})}$
answered Jan 11 at 20:28
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
The map $f$ is definitely not surjective because it misses everything in $mathbb{R}setminus{f(a_{1}),f(a_{2}),f(a_{3}),f(a_{4})}$
answered Jan 11 at 20:28
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
The map $f$ is definitely not surjective because it misses everything in $mathbb{R}setminus{f(a_{1}),f(a_{2}),f(a_{3}),f(a_{4})}$
answered Jan 11 at 20:28
pwerthpwerth
3,340417
$endgroup$
The map $f$ is definitely not surjective because it misses everything in $mathbb{R}setminus{f(a_{1}),f(a_{2}),f(a_{3}),f(a_{4})}$
answered Jan 11 at 20:28
pwerthpwerth
3,340417
answered Jan 11 at 20:28
pwerthpwerth
3,340417
answered Jan 11 at 20:28
pwerthpwerth
3,340417
answered Jan 11 at 20:28
pwerthpwerth
3,340417
3,340417
add a comment |
add a comment |
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4
$begingroup$
There are more than 4 real numbers.
$endgroup$
– Randall
Jan 11 at 20:26
1
$begingroup$
Just for general knowledge (it sounds somewhat related): It's true to say that a function $f:Arightarrow B$ for two finite sets $A,B$ with $|A|=|B|$ is injective if and only if it is surjective. However this doesn't seem to be the case in your question.
$endgroup$
– Yanko
Jan 11 at 20:30