Proof involving functions

[Jacobpm64]

Homework Statement

Prove the following:

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

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.

 

[mighty2000]

Thank you for your question. Your attempt at a solution is on the right track, but there are a few things that can be clarified and expanded upon. Firstly, when proving a statement like this, it is important to start by stating the given information clearly. In this case, you are given two functions, f and g, and you want to prove that their intersection (f ∩ g) is also a function. So your proof can start with something like: "Given two functions f : A → B and g : C → D, we want to show that their intersection f ∩ g is also a function."

Next, you correctly start by choosing an arbitrary element a ∈ A. However, you can also choose an arbitrary element c ∈ C, since the intersection of two sets A and C will contain both elements a and c. So you can say something like: "Let a ∈ A and c ∈ C be arbitrary elements."

Then, as you have done, you can use the fact that f and g are functions to show that there exist elements b ∈ B and d ∈ D such that (a, b) ∈ f and (c, d) ∈ g. However, you can make this clearer by stating it as: "Since f is a function from A to B, there exists an element b ∈ B such that (a, b) ∈ f. Similarly, since g is a function from C to D, there exists an element d ∈ D such that (c, d) ∈ g."

Next, you correctly note that the intersection of f and g will either be a function from the set containing a to the set containing b, or it will be a function from the empty set to the empty set. However, it would be better to consider both cases separately and show that in each case, the intersection is indeed a function. For example, you could start by considering the case where a = c, and show that in this case, the intersection is a function from {a} to {b}. Then, you could consider the case where a ≠ c, and show that in this case, the intersection is a function from ∅ to ∅.

Finally, you can conclude your proof by summarizing what you have shown. For example, you could say: "Therefore, the intersection f ∩ g is a function from A ∩ C to B ∩ D, as desired."