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I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?

In short, what is the benefit or reason to define a function this way? At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).

Thanks,

Moshe

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Note that the "one value" constraint is less restrictive than you might think. For example a function $f: \mathbb{R} \to \wp(\mathbb{R})$ assigns a set of real numbers to any input. –  WimC Jan 6 '13 at 16:30
    
@WimC Ermm ...but a function which maps a real to a set assigns a single value (a set) to a given argument. –  Peter Smith Jan 6 '13 at 16:37
    
@PeterSmith, I think WimC is pointing out if the codomain can be varied - and specifically to the powerset of the OP's "y-values"- then any number of y-values can be assigned while still having a function. –  alancalvitti Jan 6 '13 at 16:41
    
Of course, a function from $X$ to $\mathcal P^Y$ is the same thing as a relation between $X$ and $Y$. These are only the same thing as a multivalued function if you allow multivalued to include zero values, which is not usual. –  Rahul Jan 6 '13 at 16:56

3 Answers 3

There is an interesting relevant post by Tim Gowers on his blog http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/

It is worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".

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Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $\;x = a\;$ outputs more than one distinct value, so that $y = f(a) \in\{y_1, y_2, ..., y_k, ...\} $. We'd never be able to say what, precisely, $y$ is when $x = a$.

If $y = f(a) = b$, and $y = f(a) = c$, and $\,b\ne c$, then the value of the function $\,\,f(x) = y\,$ at $\,a\,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.

E.g. How would we define continuity of, say, a real-valued multifunction?

The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.

But you'd might like to explore the following:

See this entry on multi-valued "functions":

A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.

The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]

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Yet some texts in Complex Analysis do very cheerfully talk of many-valued functions ... E.g. Schaum's Outline of Complex Variables, 2ed (Schaum's Outline Series) by Murray Spiegel, Seymour Lipschutz, John Schiller, Dennis Spellman. This (2009 2nd edn. but first goes back to 1964: surely aims to be "middle of the road") explicitly talks about multi-valued functions (see §2.2, §2.6). –  Peter Smith Jan 6 '13 at 16:30
    
Yes, indeed. I was referring to the "strict" definition students are usually introduced to. –  amWhy Jan 6 '13 at 16:34

Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.

A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = \sqrt{1-y^2}$ and $F(x) = - \sqrt{1-y^2}$. If we had said that $F(x) = \pm \sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.

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You may wish to read about Kakutani's Fixed Point Theorem. As I mentioned in another comment this result is used in Gerard Debreu's Theory of Value. This is a short book in mathematical economics. Anyone who knows undergraduate analysis should be able to read it. I will make no claim that the theory in this book applies to the real world. –  Jay Jan 6 '13 at 23:02

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