If A = {1,2,3} and B = {w,x,y,z},then the domain of f is A and codomain is B. However what about the range?
Why is the range f=f(A)={w,x}, why cant it be {w,z}?
Edit: f ={（1,w),(2,x),(3,x)}
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Whether it is ${w,z}$ or not simply depends on which function $f$ is. And you have not told us that. – Michael Hardy Jul 14 '14 at 20:05
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Okay i made edit. – Belphegor Jul 14 '14 at 20:11
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I think the fact that you didn't know that that information was needed is the reason you didn't understand this. When you understand why the range is what it is, then you will understand the need for that information. – Michael Hardy Jul 14 '14 at 21:05
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See Range :
In mathematics, the range of a function refers to either the codomain or the image of the function, depending upon usage. Modern usage almost always uses range to mean image. The word range may eventually become obsolete.
We have that
$f : A \rightarrow B$
and the "definition" of $f$ is : $f = \{（1,w),(2,x),(3,x) \}$.
Thus the range or image of $A$ under $f$ , i.e. $f(A) \subseteq B$ is "made of" $x,w$; i.e. $f(A) = \{ w, x \}$, simply because they are the elements of $B$ in which the elements of $A$ are "mapped to" by $f$.
Mauro ALLEGRANZA
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The codomain of a function is the set into which the function is constrained to map. The range (properly called the image) of a function is the set into which the function actually maps.
An example that would be similar to your problem is $f: \Bbb{R} \to \Bbb{R}$ by $f(x) = x^2$. Although the codomain of $f(x) = x^2$ is $\Bbb{R}$, $f(x) = x^2$ truly only maps into the non-negative real numbers. So the codomain of $f(x) = x^2$ is different from the range (image).
Scott Caldwell
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$$f = \{(1,w),(2,x),(3,x)\}$$
The range is the set of things that occur as the second component of a pair. The only ones I see in that role are $w$ and $x$. That is why the range is $\{w,x\}$ and not something else.
