Question 3. Let f:X --> Y be a function. (a) Recall that for C CY, the inverse image of C is the set f-1(C) = {x E X|f(x) E C}. Let A, B CY. Prove that f-1(ANB) Cf-1(A) n f-1(B). (b) Recall that for D C X, the image of D under f is the set f(D) = {f(x) E YX E X}. Let A, B C X. Prove that f(An B) c f(A)n F(B). Find an example of a function f and sets A and B such that f(ANB) = Ø but f(A) n f(B) + Ø.

a) We want to prove that [tex]f^{-1}(A\cap B)\subset f^{-1}(A)\cap f^{-1}(B)[/tex]. Then, we can do that proving that every element of [tex]f^{-1}(A\cap B)[/tex] is an element of [tex]f^{-1}(A)\cap f^{-1}(B)[/tex] too.

Then, suppose that [tex]x\in f^{-1}(A\cap B)[/tex]. From the definition of inverse image we know that [tex]f(x)\in A\cap B[/tex], which is equivalent to [tex]f(x)\in A[/tex] and [tex]f(x)\in B[/tex]. But, as [tex] f(x) \in A [/tex] we can affirm that [tex]x\in f^{-1}(A)[/tex] and, because [tex]f(x)\in B[/tex] we have [tex]x\in f^{-1}(B)[/tex].

Therefore, [tex]x\inf^{-1}(A)\cap f^{-1}(B)[/tex].

b) We want to prove that [tex]f(A\cap B) \subset f(A)\cap f(B)[/tex]. Here we will follow the same strategy of the above exercise.

Assume that [tex]y\in f(A\cap B)[/tex]. Then, there exists [tex]x\in A\cap B[/tex] such that [tex]y=f(x)[/tex]. But, as [tex]x\in A\cap B[/tex] we know that [tex]x\in A[/tex] and [tex]x\in B[/tex]. From this, we deduce [tex]f(x)=y\in f(A)[/tex] and [tex]f(x)=y\in f(B)[/tex]. Therefore, [tex]y\in f(A)\cap f(B)[/tex].

c) Consider the constant function [tex]f(x)=1[/tex] for every real number [tex]x[/tex]. Take the sets [tex]A=(0,1)[/tex] and [tex]B=(1,2)[/tex].

Notice that [tex]A\cap B = (0,1)\cap (1,2)[/tex]=Ø, so [tex]f(A\cap B)[/tex]=Ø. But [tex]f(A) = \{1\}[/tex] and [tex]f(B) = \{1\}[/tex], so [tex]f(A)\cap f(B) =\{1\}[/tex].