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I'm having troubles proving that if $f,g \subset A \times B$ are functions, then $f\cap g$ is also a function.

Thanks for your help!

Kind regards,

Phi.

  • 0
    What have you tried? How about small sets: $A = \{a_1, a_2\}$, $B = \{b_1, b_2\}$? Here you can simply enumerate by hand all the possible functions from $A$ to $B$. – 2017-02-13

3 Answers 3

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Definition: let $h, E, F$ a sets, we define: $$h \text{ is relation if }\,\big[\forall x \in h:\exists y,\exists z: x=(y,z)\big] \\\operatorname{dom}(h):=\{x|\exists y:(x,y)\in h\} \\ \operatorname{cod}(h):=\{x|\exists y:(y,x)\in h\} \\ h \text{ is function if }\, \big[h\text{ is relation} \wedge \forall x,\forall y,\forall z: (x,y) \in h \wedge (x,z)\in h \to y=z\big] \\ h: E\to F \text{ for }\big[ h\text{ is function} \wedge \operatorname{dom}(h)= E \wedge \operatorname{cod}(h)\subseteq F\big]$$

Thereom: let be $f,g$ functions and $f \subseteq (A \times B)\supseteq g$: $$ (f \cap g) \text{ is function}$$

Proof:

  • $(f\cap g)$ is relaction because $f \subseteq (A \times B)\supseteq g$

  • let be $x,y,z$: $$ \begin{align}(x,y) \in (f \cap g) \wedge (x,z)\in (f\cap g) \leftrightarrow& \\ \big((x,y) \in f \wedge (x,y) \in g\big) \wedge \big((x,z) \in f \wedge (x,z) \in g\big) \leftrightarrow& \\ \big((x,y) \in f \wedge (x,z) \in f\big) \wedge \big( (x,y) \in g \wedge (x,z) \in g\big) \to& \\\ (y=z) \wedge (y=z) \text{ because } f \text{ and } g \text{ are by hyp. functions} \to& \\y=z \,\text{ (Q.E.D)} \end{align}$$

  • naturally $(f \cap g) \subseteq (A\times B)$, in fact $$ \begin{align} x \in (f\cap g) \leftrightarrow x \in f\wedge x \in g \to& \\ x \in (A\times B) \wedge x \in (A \times B) \,(\text{because } f \subseteq (A \times B)\supseteq g) \to&\\ x \in (A \times B) \,\text{ (Q.E.D)}\end{align}$$

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$f$ and $g$ are functions which mean that for each $a\in A$ there is a unique $b\in B$ such that $f(a)= b$. And similarly for $g$. First show that $f\cap g$ is a subset of $A\times B$ then show that it has the uniqueness property that $f$ and $g$ have. The trick is to show that for all $x$ in the domain of $f\cap g$ it is either in the domain of $f$ and the domain of $g$. Thus it will inherit the uniqueness property from $f$ and $g$.

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you know a function is a relation that satisfices $\forall a\in A,\exists!b\in B$ such that $(a,b)\in f$ or $afb$ or $f(a)=b$, depending on notation. Notice the domain of $f\cap g$ is not $A$ unless $f=g$, so for any element $(a,b)\in f\cap g$, b is the only element from $B$ related to $a$, as is related by $f$ and $g$