Question

1. Let f and g be functions with the same domain and codomain (let A be the domain and B be the codomain). Consider the following ordered triple h = (A, B, f LaTeX: \cap ∩ g) (Note: The f and g in the triple refer to the "rules" associated with the functions f and g). Prove that h is a function. Would the same thing be true if, instead of intersection, we had a union? If your answer is yes, prove it. If your answer is no, provide a counterexample.

Answer 1

Let f and g be functions with the same domain and codomain (le
... Let f and g be functions with the same domain and codomain (eet a be the domain and o be the codomain. Ars: since we restrict ourselves to elements of a, the domain of fng must be a subset of a more over for every element XEA, there in a y EB Such that (x,y) & fing. Therefore, the domain of fing is exactly A. Now we need to show that for all a EA, YEB, ZEB; if (x, y) e fng and (x,z) € ing then y: z so hence any (ory) e of ng forz) & fag . (x,y) & fag = (x,y) .Ef & (x,y) eg -> food ay a gox) = y Siniorgia) e fago fro: 1 of gr) = 2 Since, g of are functions so for every clements in their domain . they have only one image in co domain, so y: 2 . Therefore fag is a function.

fug is not always a fuckton Let A {1,2,3,4,5} B: {a,b,c,d, minip} f(x) = {a,b,c,d} g (A) = {d, minip} Also, but fin= a & goi)=m. then (1,a) & fug and com) e fuga but afm. so, fug is not a mapping or function
t A be the domain and B be the codomain.