# Proof involving functions

1. Oct 27, 2007

### Jacobpm64

1. The problem statement, all variables and given/known data
Prove the following:

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

3. The attempt at a solution

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?