3
$\begingroup$

Page 14 of Fundamentals of Computer Graphics states that if we have a function like this:

enter image description here

...the set that comes before the arrow is called the domain of the function, and the set on the right-hand side is called the target.

...The point f(a) is called the image of a, and the image of a set A (a subset of the domain) is the subset of the target that contains the images of all points in A. The image of the whole domain is called the range of the function.

Then what exactly is the difference between the range of a function and the target of a function?

$\endgroup$
1

1 Answer 1

Reset to default
8
$\begingroup$

The difference is that the range may not be the entire target. Consider the function $f(x)=x^2$ from $\{-1,0,1\}$ to $\{-1,0,1\}$; its domain and target are both $\{-1,0,1\}$, but its range is only $\{0,1\}$.

By the way, the more usual name for the target is codomain. The range is sometimes called the image.

$\endgroup$
5
  • $\begingroup$ Do you mean that the target is arbitrarily defined and can be anything? $\endgroup$
    – Pacerier
    Jul 10, 2013 at 7:59
  • $\begingroup$ @Pacerier: Anything that includes the actual range, yes. $\endgroup$
    – Brian M. Scott
    Jul 10, 2013 at 7:59
  • $\begingroup$ Hmm, then isn't the range a more useful term since the target could be well.. anything. What's the point of target anyway, when we already have range? $\endgroup$
    – Pacerier
    Jul 10, 2013 at 8:46
  • 1
    $\begingroup$ @Pacerier: They’re useful in different ways, though the value of the target only becomes fully apparent at a more advanced level. For a simple example, though, consider whether a function $f$ with domain $D$ is surjective (or onto). You can’t tell if you don’t know the target: the range doesn’t give you enough information. And there are times when it’s important to know whether a function is surjective. $\endgroup$
    – Brian M. Scott
    Jul 10, 2013 at 8:50
  • $\begingroup$ It would be fascinating to know what the downvoter imagines is wrong. $\endgroup$
    – Brian M. Scott
    Feb 11, 2022 at 23:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .