Why the more modern point of view does not admit "double-valuedness" as a property of function? In Apostol book I 2nd p.55 He says :

There was a time when mathematicians would say that y is a double-valuedfunction of x given by y=$\pm \sqrt{r^2-x^2} $. However , the more modern point of view does not admit "double-valuedness" as a property of function 

why mathematicians change their way of thinking about function ?
Edit
I have anther two questions :
1- why adding or multiply... two functions define only if they have the same domain ?
2- Function is just a set of order pairs right ? 
how can we add and multiply and divide  sets ?
thanks :)))
 A: As a short, quippy answer, because set-valued functions don't have derivatives.  What is $\frac{\mathrm{d}}{\mathrm{d}x} \pm \sqrt{r^2 - x^2}$?  Even assuming $-r \leq x \leq r$, picking $x_0 \in (-r,r)$ and picking a sequence $(x_i)_i$ with $x_i \rightarrow x_0$, every pair of $x_i$, $x_{i+1}$ has four possible values of difference quotient : two of them correspond to the two monovalent branches which we would talk about now.  The other two we expect to be diverging to $\infty$ and $-\infty$.
What about the logarithm?  It's infinite-valued.  The set of useful difference quotients is dwarfed by the horde of useless ones.  Nevertheless, multivalued functions have been studied.  The derivative is replaced with the subderivative.  But you don't really recover the information that is obtained from the derivative of single-valued functions.
For your follow-on questions (which really should have gone in a separate Question):
Consider $$f(x) + g(x)  \text{.}  $$  This says "take the value of $f$ at $x$ and add it to the value of $g$ at $x$.  There is no way to do this if $f$ and $g$ are not both defined at $x$.  That is $x$ must be in the domain of both functions for this to make sense.  One way to ensure this is to require that both functions have the same domain -- then at least, if either function is defined, both are.  Another way is to restrict the domain of the sum to the intersection of the domains of $f$ and $g$.  (This method of restriction to the intersection seems to be the general definition of the domain of a sum of functions.)
The same comments apply to $$f(x) \cdot g(x)  \text{.}  $$
A function can be modeled as a collection of ordered pairs.  But the model is not the thing.
"how can we add and multiply and divide sets"?:  We don't.  The definition of these operations shows that we do not work with the sets explicitly.  For instance, the sum of two functions is defined as
$$(f+g)(x) = f(x) + g(x)  \text{.}  $$
This says "to evaluate (the sum of $f$ and $g$) at $x$, add $f(x)$ to $g(x)$."  If you insist on working with the models of $f$ and $g$ as sets of ordered pairs:  For all $x$ in the common domain of $f$ and $g$, take the pairs $(x, f(x))$ and $(x,g(x))$ from the models of $f$ and $g$ as sets of ordered pairs, then construct the model of the function $(f+g)$ via
$$ (x,(f+g)(x)) = (x,f(x) + g(x))  \text{.}  $$
Notice that at no point did we perform arithmetic on sets.  At all times, arithmetic was performed on values.  We used arithmetic on values and other set operations to construct the model of the function $f+g$.  But this definition is clear: "$(f+g)$" does not mean to add sets together; it means to construct the function whose values are the sums of $f$ and $g$ evaluated at points of their common domain.
A: By convention, functions are always single valued. There's still "multi-valued functions," but its probably better to call them relations. Check out the wikipedia page on the category of sets and relations between them for more information. If you've reached the appropriate level of mathematical maturity, you may wish to view this as the the category structure on $\mathbf{Set}$ induced by with the powerset monad. Look up the term "Kleisli category" for more information (if you're ready.)
