# Thread: Proof involving functions

1. ## Proof involving functions

Prove the following:

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

Here's my thoughts/attempt:

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

This is all I have so far.

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

2. Originally Posted by Jacobpm64
Prove the following:
If $\displaystyle f : A \rightarrow B$ and $\displaystyle g : C \rightarrow D$, then $\displaystyle f \cap g : A \cap C \rightarrow B \cap D$.
As stated the proposition is false.
It may be that $\displaystyle f \cap g = \emptyset$.
Then what?
Please check the exact wording of the question

3. my homework just says Prove the following theorem.

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

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

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

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

5. I'm convinced now.

Thanks.