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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!

frank
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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". – Alex Provost Feb 27 '16 at 02:33
  • My question is, why in first calculus courses, one is told that a function is single-valued by definition? – frank Feb 27 '16 at 02:39
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    Because that's how we define a function in all of mathematics! – Alex Provost Feb 27 '16 at 02:55
  • In particular, in topology. Which multivalued function from topology do you have in mind? – Alex Provost Feb 27 '16 at 03:15
  • 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 $x\mapsto [x]$ from a general topological space X onto X/R (where R is a certain equivalence defined over $X$ and $X\R$ is the associated quotient space) is in general not single-valued. – frank Feb 27 '16 at 03:17
  • Yes, the quotient map $X \to X/R$ is single-valued! It associated to each element $x \in X$ the unique equivalence class $[x] \in X/R$. – Alex Provost Feb 27 '16 at 03:19
  • Exactly, so the point is that it depends on what we mean as unique. Such an equivalence class can be clearly a set, though we look at that as a point due to the definition of $X\setminus R$. I see now, so you meant that multi-valued functions are called so because we usually study those in the Euclidean space. In such a space, single-valued means mapping a point to another point. Is that correct? – frank Feb 27 '16 at 03:27
  • Multivalued functions don't really have anything special to do with Euclidean spaces. They're simply objects that are more general than functions. And relations are even more general than multivalued functions. Also, being multivalued or not has nothing to do with being set-valued or not. Like I said, in a sense every mathematical object you consider is a set (if you take some theory of sets as your foundations of mathematics). – Alex Provost Feb 27 '16 at 04:06
  • As an aside. Each multivalued function from $A$ to $B$ may be considered as a single valued function from $A$ to the power set of $B$ and vice versa. Both objects/concepts are equivalent. – user251257 Feb 27 '16 at 07:19

2 Answers2

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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 and only 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.

Taylor Rendon
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Alex Provost
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  • Still it's not clear...to me set-valued and multi-valued functions are the same object. Thank you for replying. – frank Mar 04 '16 at 02:19
  • @frank I don't understand your concern. E.g. $R_2$ defines a standard (single-valued) function which is "set-valued". – Alex Provost Mar 04 '16 at 02:22
  • 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 :) – frank Mar 04 '16 at 02:29
  • @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". – Alex Provost Mar 04 '16 at 03:27
  • 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. – IV_ Apr 22 '17 at 22:44
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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.

Allawonder
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