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|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.

Answer 1

Answer:

Step-by-step 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 $f(A)\cap f(B) =\{1\}$.