Why does g^{-1}(A∪B)=g^{-1}(A)∪g^{-1}(B) instead of intersection?

[mtayab1994]

Homework Statement

Let g be a function from ℝ to ℝ and for all subsets A and B of R.

Homework Equations

Prove that:
$$g^{-1}(A\cup B)=g^{-1}(A)\cap g^{-1}(B)$$
and
$$g^{-1}(A\cap B)=g^{-1}(A)\cup g^{-1}(B)$$

The Attempt at a Solution

Our teacher gave us this problem but I think it's wrong because I was easily able to prove that:
$$g^{-1}(A\cup B)=g^{-1}(A)\cup g^{-1}(B)$$
and
$$g^{-1}(A\cap B)=g^{-1}(A)\cap g^{-1}(B)$$
but when i try to solve what he gave us this is all I can do:

$$x\in g^{-1}(A\cup B)\Longleftrightarrow g(x)\in A\cup B\Longleftrightarrow g(x)\in Aorg(x)\in B\Longleftrightarrow x\in g^{-1}(A)\cup g^{-1}(B)$$

and the same goes for the second one. Can anyone tell me where I'm going wrong?

 

[pasmith]

Homework Statement

Let g be a function from ℝ to ℝ and for all subsets A and B of R.

Homework Equations

Prove that:
$$g^{-1}(A\cup B)=g^{-1}(A)\cap g^{-1}(B)$$
and
$$g^{-1}(A\cap B)=g^{-1}(A)\cup g^{-1}(B)$$

These are wrong. A counterexample to both is the zero function $g(x) \equiv 0$ with $A = \{0\}$ and $B = \{1\}$.

The Attempt at a Solution

Our teacher gave us this problem but I think it's wrong because I was easily able to prove that:
$$g^{-1}(A\cup B)=g^{-1}(A)\cup g^{-1}(B)$$
and
$$g^{-1}(A\cap B)=g^{-1}(A)\cap g^{-1}(B)$$

These are correct.