Question

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. 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.

Answer 1

Explanation:

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

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

Therefore,
`x\inf^(-1)(A)\cap f^(-1)(B)`.

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

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

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

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