Essential requirements for the definition of a function.

The book I'm working on states that a function f: A -> B is defined under the following conditions:

1. Every x element of A associated with some y in B, i.e., a function is defined only when the domain is entirely "used up". The set B may not be entirely "used up".
2. The function may associate more than one x to the same y.
3. No element A should have more than one image in B

Now, I have three question listed as follows:

1. What do they mean by a function being "defined"? Is there "undefined" functions?
2. Is there a reason why the set A has to be entirely "used up" and not B, or is it because that's how mathematicians define it?
3. If we have a function that relates two sets, for example A and B, the domain is A and the codomain B, but because the the set B may not be entirely "used up", then there will be a subset of the codomain, let's say, C that will represent the range of the function. Right? If so, does this mean that when we restrict a function codomain, the nonerestricted part of the codomaim will become the range?

1 Answer

Q1: What do they mean by a function being defined? They just mean that is satisfies the definition of a function. i.e., Every element of the domain must be mapped somewhere, and no single element of the domain can be mapped to two different elements.

Let $A$ be the set $\{a,b\}$, and let $B$ be the set $\{1,2,3\}$. Consider the following$$f:A→ B$$which satisfies $f(a)=1, f(b)=2$. Every element of $A$ is "used up" (we told $f$ where to send $a$ and where to send $b$), and there is no single element of $A$ that is sent to two different places. So, $f$ is a function, i.e., it is defined. In order for something to be a function, it doesn't matter how many elements of $B$ are "hit" by elements of $A$ after applying the function. In this case, $3$ is not hit by any element of $A$ after applying the function, but that's okay, because every element of $A$ is mapped somewhere.

Now, with $A$ and $B$ the same as above, if we now consider$$g:A→ B$$given by $g(a)=1,g(a)=2,g(b)=3$, note that the element $a$ is mapped to two different elements of $B$; a is mapped to 1, and to 2. This is not a function, because one single element of $A$ is mapped to two different elements of the codomain. This doesn't mean it's not useful to consider maps such as this, but we just don't call them functions. So, this 'function' (not really a function) is not defined.

Q2: Is there a reason $A$ has to be entirely used up? I think the silly answer is, yes, that's just the way it's defined. If a function $f:A→ B$ only used up some subset $A'⊂ A$, then we can just toss out the extra elements of A that aren't used up, and consider the same function $f:A'→ B$.

Q3 Your statement is correct, that in general, since $B$ is not "used up" by the function, there is some subset of $B$ which will be the range of the function. When we "restrict" (when in reference to the codomain, we say 'constrict' or 'co-restrict') the codomain we are removing elements of the codomain that are not used up by the function. So, yes, the "unrestricted" part, aka the part that we didn't touch, will become the range only when we remove all elements that are not used up by the function. Usually we constrict a function fully, so that the new codomain is exactly the range.