How to prove the intersection of functions is a function

Question

My question reads: Prove that, if f and g are functions, then f ∩ g is a function by showing that f∩g=g|A.

Answer 1

OK, $f \cap g = \{ (x,y) | (x,y) \in f \land (x,y)\in g\}$. Let's first show that this is the same set as $g|A = \{(x,y) \in g | x \in A\}$.

OK, take some $(x,y) \in f\cap g$. Then $(x,y) \in g$ and $(x,y) \in f$. Since $f$ and $g$ are functions, there is exactly one $y$ such that $(x,y)\in f$ and same for $g$ .. Which is why we can refer to this $y$ as $f(x)$ and $g(x)$ respectively. But since it is the same $y$, we have that $f(x) = g(x)$, so $x \in A$. so, $(x,y) \in g$ and $x\in A$, so $(x,y) \in g|A$.

Ok, now let's take some $(x,y) \in g|A$. This means $(x,y)\in g$ and $x \in A$. Since $x\in A$, we have that $f(x)=g(x)$. but since $(x,y)\in g$ that means $g(x)=f(x)=y$. So, $(x,y)\in f$. So, $(x,y)\in f $ and $(x,y)\in g$, so $(x,y)\in f\cap g$.

So, $f \cap g = g|A$.

The domain of $f \cap g$ is a subset of $A$ (that is, since we restrict ourselves to elements of $A$, no elements from outside A can possibly in the domain of $f \cap g$). Moreover, for every element $x \in A$ there is a $y$ (namely $y = g(x)$ such that $(x,y) \in f \cap g$. So, the domain of $f \cap g$ is exactly $A$, i.e. $dom(f \cap g)=A$.

OK, now we need to show that for all $x \in A,y \in B,z\in B$: if $(x,y) \in f \cap g$ and $(x,z) \in f \cap g$ then $y = z$ (you actually don't have that defined quite right in your comment)

OK, so take any such $(x,y) \in f \cap g$ and $(x,z) \in f \cap g$. By definition of $f \cap g$, we therefore have $(x,y) \in g$ and $(x,z) \in g$. And since $g$ is a function, we therefore have that $y=z$, as desired.

So yes, $f \cap g$ is a function.