Equality of functions

1. Aug 13, 2008

evagelos

given the definition of a function can the following be proved as a theorem?

.......for all f,g f=g iff Df=Dg and for all x ,xεDf -------.f(x)=g(x).............

........where Df is the domain of f,Dg is the domain of g..........................

2. Aug 14, 2008

CompuChip

It seems pretty obvious to me that it is true. To prove it:

- Assume that f = g. By considering only the first coordinate of each pair, you can show that $Df \subset Dg \text{ and } Dg \subset Df$, then take any x and show that the function values are equal.

- Next assume the right hand side, let $(x, y) \in f$ and show that it is in g, then show the converse inclusion.

All looks pretty trivial... try it.

3. Aug 14, 2008

evagelos

thanks,but although your proof is a skeleton of the supposed to be proof and i am not so sure if this can be done or not, it means that the definition of equality of functions is wrong since there is a theorem to support the equality

4. Aug 14, 2008

CompuChip

OK so let's first agree on a definition of function. A function is a set F of ordered pairs (x, y) with the x from a set X, and the y from a set Y. Is X allowed to be bigger than strictly the set $\{ x \in X | \exists y: (x, y) \in F \}$? Similarly for Y? Is the function allowed to be multi-valued (e.g. there may exist different y, y' such that (x, y) and (x, y') are both in F)?

5. Aug 14, 2008

evagelos

My definition of a function is the following:
f: A----->B iff..........1) f is a subset of AxB and.......
................................2) for all xεA there exists a unique yεB such that (x,y)belongs to f

6. Aug 14, 2008

CompuChip

So then if you have two functions
f, g: A --> B
you automatically have that their domains agree (since the domain is all of A by definition and A = A ) and since the subsets of AxB are the same, you can take any element from A and the function values will agree automatically. Conversely, if the domains are the same and the function values for each element from the domain are the same, then you immediately get that f = g (as relations) from f = g (as functions).

So the "definition" of f = g (as functions) as you gave in the first post is completely natural and is equivalent to f = g (as relations). So it is right that in any analysis course, "the domains are the same and all the function values agree" is the definition used (because it actually follows from a set-theoretical definition of function, though in analysis courses functions are usually not rigorously defined other than "a prescription that maps elements to other elements").

7. Aug 18, 2008

evagelos

no, it is not f,g:A--------->B but f: Df--------->.Rf and g:Dg---------->Rg

and we must prove that Df=Dg

8. Aug 19, 2008

CompuChip

In your definition from post #5, you assign a value to each element from A, so Df = Dg = A.