# Proof involving functions

1. Oct 27, 2007

### 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?

2. Oct 27, 2007

### HallsofIvy

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

3. Oct 27, 2007

### Jacobpm64

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

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

4. Oct 27, 2007

### 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).

5. Oct 27, 2007

### 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.

6. Oct 27, 2007

### Hurkyl

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

7. Oct 27, 2007

### ZioX

Well there you go: it's not true.

Unions are functions when the domains are disjoint.

8. Oct 27, 2007

### Jacobpm64

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

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

Scares me.