Proof involving functions

**Jacobpm64** · October 27th 2007, 09:38 AM

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.

**Plato** · October 27th 2007, 10:29 AM

Originally Posted by Jacobpm64

Prove the following:
If $f : A \rightarrow B$ and $g : C \rightarrow D$ , then $f \cap g : A \cap C \rightarrow B \cap D$ .

As stated the proposition is false.
It may be that $f \cap g = \emptyset$ .
Then what?
Please check the exact wording of the question

**Jacobpm64** · October 27th 2007, 10:35 AM

my homework just says Prove the following theorem.

Theorem 1
If $f : A \rightarrow B$ and $g : C \rightarrow D$ , then $f \cap g : A \cap C \rightarrow B \cap D$ .

Isn't $f \cap g$ still a function even if it is the empty set? In this case the definition of function would be vacuously true, right?

**Plato** · October 27th 2007, 11:55 AM

As stated the proposition is still false.
Consider this example $A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}$
$f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}$
$g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}$

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.

**Jacobpm64** · October 27th 2007, 12:04 PM

I'm convinced now.

Thanks.

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