On the meaning of set-valued mappings
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here one question that may look stupid. Why in general one insists on naming in a different way functions and set-valued functions just because one is single valued and the other is not? I mean, from topology, we define as function an object that maps one topological space $X$ into another one $Y$, we never require this mapping to be single-valued! Thanks!
functions
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show 4 more comments
$begingroup$
here one question that may look stupid. Why in general one insists on naming in a different way functions and set-valued functions just because one is single valued and the other is not? I mean, from topology, we define as function an object that maps one topological space $X$ into another one $Y$, we never require this mapping to be single-valued! Thanks!
functions
$endgroup$
$begingroup$
What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
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– Alex Provost
Feb 27 '16 at 2:33
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My question is, why in first calculus courses, one is told that a function is single-valued by definition?
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– frank
Feb 27 '16 at 2:39
1
$begingroup$
Because that's how we define a function in all of mathematics!
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– Alex Provost
Feb 27 '16 at 2:55
$begingroup$
In particular, in topology. Which multivalued function from topology do you have in mind?
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– Alex Provost
Feb 27 '16 at 3:15
$begingroup$
That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
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– frank
Feb 27 '16 at 3:17
|
show 4 more comments
$begingroup$
here one question that may look stupid. Why in general one insists on naming in a different way functions and set-valued functions just because one is single valued and the other is not? I mean, from topology, we define as function an object that maps one topological space $X$ into another one $Y$, we never require this mapping to be single-valued! Thanks!
functions
$endgroup$
here one question that may look stupid. Why in general one insists on naming in a different way functions and set-valued functions just because one is single valued and the other is not? I mean, from topology, we define as function an object that maps one topological space $X$ into another one $Y$, we never require this mapping to be single-valued! Thanks!
functions
functions
asked Feb 27 '16 at 2:28
frankfrank
596
596
$begingroup$
What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
$endgroup$
– Alex Provost
Feb 27 '16 at 2:33
$begingroup$
My question is, why in first calculus courses, one is told that a function is single-valued by definition?
$endgroup$
– frank
Feb 27 '16 at 2:39
1
$begingroup$
Because that's how we define a function in all of mathematics!
$endgroup$
– Alex Provost
Feb 27 '16 at 2:55
$begingroup$
In particular, in topology. Which multivalued function from topology do you have in mind?
$endgroup$
– Alex Provost
Feb 27 '16 at 3:15
$begingroup$
That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
$endgroup$
– frank
Feb 27 '16 at 3:17
|
show 4 more comments
$begingroup$
What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
$endgroup$
– Alex Provost
Feb 27 '16 at 2:33
$begingroup$
My question is, why in first calculus courses, one is told that a function is single-valued by definition?
$endgroup$
– frank
Feb 27 '16 at 2:39
1
$begingroup$
Because that's how we define a function in all of mathematics!
$endgroup$
– Alex Provost
Feb 27 '16 at 2:55
$begingroup$
In particular, in topology. Which multivalued function from topology do you have in mind?
$endgroup$
– Alex Provost
Feb 27 '16 at 3:15
$begingroup$
That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
$endgroup$
– frank
Feb 27 '16 at 3:17
$begingroup$
What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
$endgroup$
– Alex Provost
Feb 27 '16 at 2:33
$begingroup$
What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
$endgroup$
– Alex Provost
Feb 27 '16 at 2:33
$begingroup$
My question is, why in first calculus courses, one is told that a function is single-valued by definition?
$endgroup$
– frank
Feb 27 '16 at 2:39
$begingroup$
My question is, why in first calculus courses, one is told that a function is single-valued by definition?
$endgroup$
– frank
Feb 27 '16 at 2:39
1
1
$begingroup$
Because that's how we define a function in all of mathematics!
$endgroup$
– Alex Provost
Feb 27 '16 at 2:55
$begingroup$
Because that's how we define a function in all of mathematics!
$endgroup$
– Alex Provost
Feb 27 '16 at 2:55
$begingroup$
In particular, in topology. Which multivalued function from topology do you have in mind?
$endgroup$
– Alex Provost
Feb 27 '16 at 3:15
$begingroup$
In particular, in topology. Which multivalued function from topology do you have in mind?
$endgroup$
– Alex Provost
Feb 27 '16 at 3:15
$begingroup$
That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
$endgroup$
– frank
Feb 27 '16 at 3:17
$begingroup$
That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
$endgroup$
– frank
Feb 27 '16 at 3:17
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I believe we have lifted some of your confusion in the comments, but let me illustrate how being "set-valued" and being "multivalued" have nothing in common. Let $A = {1}$ be a one-point set and let $B = {1,A,{A}}$. We are going to define four relations from $A$ to $B$. By definition these are subsets of the three-element set $A times B$, and $a in A$ is related to $b in B$ if $(a,b)$ belongs to the relation.
Let $R_1 = {(1,1)}$. This is just an ordinary function: the input $1$ has the unique output $1$. Moreover it isn't really "set-valued" if we consider $1$ to be an object that is somehow not really a set because we haven't specified what it contains.
Let $R_2 = {(1,A)}$. Again this is just an ordinary function: the input $1$ yields the output $A$ (a set). This is a "set-valued function".
Let $R_3 = {(1,1),(1,A)}$. This is a multivalued function because the input $1$ has precisely two outputs, namely $1$ and $A$. It isn't precisely set-valued because you don't consider $1$ to be a set.
Let $R_4 = {(1,A),(1,{A})}$. This is a "set-valued" multifunction.
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$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
add a comment |
$begingroup$
Your question, as it stands, is confusing, but from your discussion with others in the comment sections, I think the problem is that you are yet to understand what the word function means in modern mathematics. In particular, I think you are imposing additional constraints on the usual definition, namely that you think a function must be one-to-one. But this is not necessary at all. Many functions are not injective (that's synonymous to one-to-one), and an example is the one you plucked from topology in your discussion in the comment under the OP.
So a function need not map different points to different images. There is absolutely no need for that in general. All that we require is that the same point must have no less and no more than one image.
In the language of sets, a function takes each point in a set (called the domain) to exactly one image point in a set (called the codomain). This does not mean two points in the domain cannot have the same image (we make no such restriction), so long as they both have just that one point as image.
I hope this clarifies some things. But I think you should read more about functions and really understand them. A definition means just what it says, no more or less.
PS. I thought I should give an example. Note that what we call a constant function is any function that takes all the points in the domain to the same one point in the codomain. This is a function because every point of the domain has precisely one image, even though in this case they all have the same image.
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
I believe we have lifted some of your confusion in the comments, but let me illustrate how being "set-valued" and being "multivalued" have nothing in common. Let $A = {1}$ be a one-point set and let $B = {1,A,{A}}$. We are going to define four relations from $A$ to $B$. By definition these are subsets of the three-element set $A times B$, and $a in A$ is related to $b in B$ if $(a,b)$ belongs to the relation.
Let $R_1 = {(1,1)}$. This is just an ordinary function: the input $1$ has the unique output $1$. Moreover it isn't really "set-valued" if we consider $1$ to be an object that is somehow not really a set because we haven't specified what it contains.
Let $R_2 = {(1,A)}$. Again this is just an ordinary function: the input $1$ yields the output $A$ (a set). This is a "set-valued function".
Let $R_3 = {(1,1),(1,A)}$. This is a multivalued function because the input $1$ has precisely two outputs, namely $1$ and $A$. It isn't precisely set-valued because you don't consider $1$ to be a set.
Let $R_4 = {(1,A),(1,{A})}$. This is a "set-valued" multifunction.
$endgroup$
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
add a comment |
$begingroup$
I believe we have lifted some of your confusion in the comments, but let me illustrate how being "set-valued" and being "multivalued" have nothing in common. Let $A = {1}$ be a one-point set and let $B = {1,A,{A}}$. We are going to define four relations from $A$ to $B$. By definition these are subsets of the three-element set $A times B$, and $a in A$ is related to $b in B$ if $(a,b)$ belongs to the relation.
Let $R_1 = {(1,1)}$. This is just an ordinary function: the input $1$ has the unique output $1$. Moreover it isn't really "set-valued" if we consider $1$ to be an object that is somehow not really a set because we haven't specified what it contains.
Let $R_2 = {(1,A)}$. Again this is just an ordinary function: the input $1$ yields the output $A$ (a set). This is a "set-valued function".
Let $R_3 = {(1,1),(1,A)}$. This is a multivalued function because the input $1$ has precisely two outputs, namely $1$ and $A$. It isn't precisely set-valued because you don't consider $1$ to be a set.
Let $R_4 = {(1,A),(1,{A})}$. This is a "set-valued" multifunction.
$endgroup$
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
add a comment |
$begingroup$
I believe we have lifted some of your confusion in the comments, but let me illustrate how being "set-valued" and being "multivalued" have nothing in common. Let $A = {1}$ be a one-point set and let $B = {1,A,{A}}$. We are going to define four relations from $A$ to $B$. By definition these are subsets of the three-element set $A times B$, and $a in A$ is related to $b in B$ if $(a,b)$ belongs to the relation.
Let $R_1 = {(1,1)}$. This is just an ordinary function: the input $1$ has the unique output $1$. Moreover it isn't really "set-valued" if we consider $1$ to be an object that is somehow not really a set because we haven't specified what it contains.
Let $R_2 = {(1,A)}$. Again this is just an ordinary function: the input $1$ yields the output $A$ (a set). This is a "set-valued function".
Let $R_3 = {(1,1),(1,A)}$. This is a multivalued function because the input $1$ has precisely two outputs, namely $1$ and $A$. It isn't precisely set-valued because you don't consider $1$ to be a set.
Let $R_4 = {(1,A),(1,{A})}$. This is a "set-valued" multifunction.
$endgroup$
I believe we have lifted some of your confusion in the comments, but let me illustrate how being "set-valued" and being "multivalued" have nothing in common. Let $A = {1}$ be a one-point set and let $B = {1,A,{A}}$. We are going to define four relations from $A$ to $B$. By definition these are subsets of the three-element set $A times B$, and $a in A$ is related to $b in B$ if $(a,b)$ belongs to the relation.
Let $R_1 = {(1,1)}$. This is just an ordinary function: the input $1$ has the unique output $1$. Moreover it isn't really "set-valued" if we consider $1$ to be an object that is somehow not really a set because we haven't specified what it contains.
Let $R_2 = {(1,A)}$. Again this is just an ordinary function: the input $1$ yields the output $A$ (a set). This is a "set-valued function".
Let $R_3 = {(1,1),(1,A)}$. This is a multivalued function because the input $1$ has precisely two outputs, namely $1$ and $A$. It isn't precisely set-valued because you don't consider $1$ to be a set.
Let $R_4 = {(1,A),(1,{A})}$. This is a "set-valued" multifunction.
answered Feb 27 '16 at 4:57
Alex ProvostAlex Provost
15.5k22350
15.5k22350
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
add a comment |
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying.
$endgroup$
– frank
Mar 4 '16 at 2:19
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
@frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued".
$endgroup$
– Alex Provost
Mar 4 '16 at 2:22
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
Exactly the point is that I'm not getting why one couldn't call valued functions simply functions. It's just a matter of giving the name to things :)
$endgroup$
– frank
Mar 4 '16 at 2:29
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
@frank Well, generally single-valued functions are much more common and more useful than multi-valued functions in most fields of mathematics, hence the convention to call the former simply "functions".
$endgroup$
– Alex Provost
Mar 4 '16 at 3:27
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
$begingroup$
A function is one-to-one or many-to-one. A set-valued function is simply a function whose values are sets. A multifunction is multi-valued. Therefore it is not a function. "Multifunction" is a misnomer. A multifunction is no function, it is a special kind of relations.
$endgroup$
– IV_
Apr 22 '17 at 22:44
add a comment |
$begingroup$
Your question, as it stands, is confusing, but from your discussion with others in the comment sections, I think the problem is that you are yet to understand what the word function means in modern mathematics. In particular, I think you are imposing additional constraints on the usual definition, namely that you think a function must be one-to-one. But this is not necessary at all. Many functions are not injective (that's synonymous to one-to-one), and an example is the one you plucked from topology in your discussion in the comment under the OP.
So a function need not map different points to different images. There is absolutely no need for that in general. All that we require is that the same point must have no less and no more than one image.
In the language of sets, a function takes each point in a set (called the domain) to exactly one image point in a set (called the codomain). This does not mean two points in the domain cannot have the same image (we make no such restriction), so long as they both have just that one point as image.
I hope this clarifies some things. But I think you should read more about functions and really understand them. A definition means just what it says, no more or less.
PS. I thought I should give an example. Note that what we call a constant function is any function that takes all the points in the domain to the same one point in the codomain. This is a function because every point of the domain has precisely one image, even though in this case they all have the same image.
$endgroup$
add a comment |
$begingroup$
Your question, as it stands, is confusing, but from your discussion with others in the comment sections, I think the problem is that you are yet to understand what the word function means in modern mathematics. In particular, I think you are imposing additional constraints on the usual definition, namely that you think a function must be one-to-one. But this is not necessary at all. Many functions are not injective (that's synonymous to one-to-one), and an example is the one you plucked from topology in your discussion in the comment under the OP.
So a function need not map different points to different images. There is absolutely no need for that in general. All that we require is that the same point must have no less and no more than one image.
In the language of sets, a function takes each point in a set (called the domain) to exactly one image point in a set (called the codomain). This does not mean two points in the domain cannot have the same image (we make no such restriction), so long as they both have just that one point as image.
I hope this clarifies some things. But I think you should read more about functions and really understand them. A definition means just what it says, no more or less.
PS. I thought I should give an example. Note that what we call a constant function is any function that takes all the points in the domain to the same one point in the codomain. This is a function because every point of the domain has precisely one image, even though in this case they all have the same image.
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add a comment |
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Your question, as it stands, is confusing, but from your discussion with others in the comment sections, I think the problem is that you are yet to understand what the word function means in modern mathematics. In particular, I think you are imposing additional constraints on the usual definition, namely that you think a function must be one-to-one. But this is not necessary at all. Many functions are not injective (that's synonymous to one-to-one), and an example is the one you plucked from topology in your discussion in the comment under the OP.
So a function need not map different points to different images. There is absolutely no need for that in general. All that we require is that the same point must have no less and no more than one image.
In the language of sets, a function takes each point in a set (called the domain) to exactly one image point in a set (called the codomain). This does not mean two points in the domain cannot have the same image (we make no such restriction), so long as they both have just that one point as image.
I hope this clarifies some things. But I think you should read more about functions and really understand them. A definition means just what it says, no more or less.
PS. I thought I should give an example. Note that what we call a constant function is any function that takes all the points in the domain to the same one point in the codomain. This is a function because every point of the domain has precisely one image, even though in this case they all have the same image.
$endgroup$
Your question, as it stands, is confusing, but from your discussion with others in the comment sections, I think the problem is that you are yet to understand what the word function means in modern mathematics. In particular, I think you are imposing additional constraints on the usual definition, namely that you think a function must be one-to-one. But this is not necessary at all. Many functions are not injective (that's synonymous to one-to-one), and an example is the one you plucked from topology in your discussion in the comment under the OP.
So a function need not map different points to different images. There is absolutely no need for that in general. All that we require is that the same point must have no less and no more than one image.
In the language of sets, a function takes each point in a set (called the domain) to exactly one image point in a set (called the codomain). This does not mean two points in the domain cannot have the same image (we make no such restriction), so long as they both have just that one point as image.
I hope this clarifies some things. But I think you should read more about functions and really understand them. A definition means just what it says, no more or less.
PS. I thought I should give an example. Note that what we call a constant function is any function that takes all the points in the domain to the same one point in the codomain. This is a function because every point of the domain has precisely one image, even though in this case they all have the same image.
edited Jun 25 '18 at 12:57
answered Jun 25 '18 at 12:48
AllawonderAllawonder
1,970616
1,970616
add a comment |
add a comment |
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What do you mean? No one makes distinctions between "functions" and "set-valued functions". If your foundations are based on set theory then everything is a set anyway. Are you asking about multivalued functions? By default when mathematicians speak of "functions" they mean "single-valued functions".
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– Alex Provost
Feb 27 '16 at 2:33
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My question is, why in first calculus courses, one is told that a function is single-valued by definition?
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– frank
Feb 27 '16 at 2:39
1
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Because that's how we define a function in all of mathematics!
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– Alex Provost
Feb 27 '16 at 2:55
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In particular, in topology. Which multivalued function from topology do you have in mind?
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– Alex Provost
Feb 27 '16 at 3:15
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That's the point! If this is true (unless I'm missing the point), then any function $f$ from a topological to another has to be single-valued, while there is plenty of function that are not so. For instance, the function $xmapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $XR$ is the associated quotient space) is in general not single-valued.
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– frank
Feb 27 '16 at 3:17