Proving the Intersection of Functions

[Jacobpm64]

Prove the following:

If f : A \rightarrow B and g : C \rightarrow D, then f \cap g : A \cap C \rightarrow B \cap D.

Here's my thoughts/attempt:

Proof:
Let A, B, C, and D be sets. Assume f : A \rightarrow B and g : C \rightarrow D. Let a \in A. Since f is a function from A to B, there is some y \in B such that (a, y) \in f. Let b \in B be such an element, that is, let b \in B such that (a,b) \in f. Let c \in C. Since g is a function from C to D, there is some z \in D such that (c, z) \in g. Let d \in D be such an element, that is, let d \in D such that (c,d) \in g.



This is all I have so far.

Would I have to break it into cases where a = c and a \not= c? If a = c, A \cap C contains an element, but if a \not= c, A \cap C is empty since a and c were arbitrary. The same argument holds for B \cap D. So, taking these things into account, f \cap g is either a function from the set containing a to the set containing b, or its a function from the empty set to the empty set.

Does this make any sense, is it necessary, and how should I write it in my proof?

Thanks in advance.

 

[HallsofIvy]

What, exactly, is your definition of f \cap g?

 

[Jacobpm64]

I'm guessing just the normal definition of intersection of sets.

All the ordered pairs that are common to both f and g.

 

[ZioX]

Just talk in terms of set theoretics. f is a set, g is a set. Show that the intersection of f and g defines a new relation on (A intersect C)x(B intersect D) that satisfies the definition of a function. (for every x in A intersect C there is some y in B intersect D such that (x,y) in the relation and that for any x in A intersect C this y is unique.)

What happens if we take unions? Do we still get a new function?

Also, no one really talks about functions this way (intersections and unions).

 

[Jacobpm64]

Well, one of our earlier assignments was to disprove the case when we took unions.

So I know that you don't get a function when you take f union g.



I'm still not convinced that the intersection claim is correct though.

Earlier, on another forum, someone came up with the counterexample:
A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}
f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}
g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}

But f \cap g = \left\{ {(2,r)} \right\} while A \cap C = \left\{ {2,4} \right\} clearly f \cap g:A \cap C \not{\mapsto} B \cap D
There is no mapping for the term 4.

 

[Hurkyl]

Also, no one really talks about functions this way (intersections and unions).

Unless you're working in an algebra of relations, in which case the operation is fairly natural.

 

[ZioX]

Well there you go: it's not true.

Unions are functions when the domains are disjoint.

 

[Jacobpm64]

i'm always afraid to write a "this is wrong"

on a homework assignment that says "prove the following theorem"

Scares me.