Not sure how to proceed: If $f: B \cup C \rightarrow A$ is a function, then $f = f_{[B]} \cup f_{[C]}$

Question

I'm using Charles Pinter's A Book of Set Theory and I'm given the following theorem to prove.

Theorem 2.15: If $f: B \cup C \rightarrow A$ is a function, then $f = f_{[B]} \cup f_{[C]}$

The book states that the proof is simple, so I think that I'm missing something fundamental. Right now I think my best bet to prove this is to work with the conditions for a function (listed below). But I'm not too sure how to proceed, as I haven't seen a similar proof to imitate. I'm particularly confused about where exactly I should break up the function $f$ into the union of $f_{[B]}$ and $f_{[C]}$, respectively.

Could anyone point out what I may be missing in the proof, or some other guidance?

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We defined a function as follows: Let $A$ and $B$ be classes and let $f$ be a graph. Then $f: A \rightarrow B$ is a function iff

1. $\forall x \in A$, $\exists y \in B$ such that $(x,y) \in f$, and the image of $x$ is unique.
2. domain $f = A$, and
3. range $f \subseteq B$.

Attempted Proof/Outline

Assume that $f: B \cup C \rightarrow A$ is a function. Now,

(1) $x \in$ dom $f$ $\Rightarrow$ $\exists y \ni$ $(x,y) \in f$

(2) $\Rightarrow$ $(x,y) \in (B \cup C) \times A$

(3) $\Rightarrow$ $x \in (B \cup C)$ $\wedge$ $y \in A$

(4) $\Rightarrow$ ($x \in B$ $\vee$ $x \in C$) $\wedge$ $y \in A$

(5) After distributing $\Rightarrow$ ($x \in B \wedge y \in A$) $\vee$ ($x \in C \wedge y \in A$)

(6) Returning to Cartesian Product $\Rightarrow$ $(x,y) \in [(B \times A) \cup (C \times A)]$

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Right here it looks like I have the function split into a union, which I want, since we defined a function to simply be a graph. The next step would be showing the converse, namely, $A \subseteq$ dom $f$. If the above is correct, I have a general idea how to proceed.

Answer 1

You're heading in the wrong direction, I'm afraid. Your approach will help you show that $\operatorname{dom} f=(\operatorname{dom} f_{[B]})\cup(\operatorname{dom} f_{[C]}),$ but that isn't what you're trying to show. Instead, you wish to show two facts: $$f\subseteq f_{[B]}\cup f_{[C]}$$ and $$f\supseteq f_{[B]}\cup f_{[C]}.$$

On the one hand, take any $\langle x,y\rangle\in f.$ Clearly, $$x\in\operatorname{dom} f=B\cup C=(\operatorname{dom} f_{[B]})\cup(\operatorname{dom} f_{[C]}),$$ so what can you conclude by definition of restriction?

The reverse inclusion is similar.

Answer 2

First your definition need be changed to:

$f: A \rightarrow B$ is a function iff

1. $\forall x \in A$, $\exists y \in B,\:y$ is unique such that $(x,y) \in f$.
2. $\mathcal{D}(f) = A$, and
3. $\mathcal{R}(f)\subseteq B$.

where $\mathcal{D}(f)$ is the domain of $f$, and $\mathcal{R}(f)$ the range of $f$.

Denote $f|_B^A=f: B \rightarrow A$ as $f$ under domain $B$ and range $A$. Then $$ f|_B^A=f\cap (B\times A) $$ where $A\subset \mathcal{R}(f)$, the range of $f$.

Next \begin{align} f|_{B\cup C}^A&=f\cap (B\cup C\times A) \\ &=f\cap (B\times A\cup C\times A) \\ &=(f\cap (B\times A))\cup (f\cap (C\times A)) \\ &=f|_{B}^A\cup f|_{C}^A \end{align}