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Here is the question:

Let X and Y be sets, and let f : X → Y be a function defined in X with values in Y. Prove, that if A, B ⊂ Y, then

f^−1 (A \ B) = f^-1 (A) \ f^−1 (B).

My attempt to solve the problem:

I need to show that:f^−1 (A \ B) ⊆ f^-1 (A) \ f^−1 (B) and f^-1 (A) \ f^−1 (B) ⊆ f^−1 (A \ B)

if A⊂Y and B⊂Y => A\B⊂Y => {x∈X: f(x)∈A\B} => f(x)∈A and f(x)∉B

if A⊂Y => {m∈X: f(m)∈A} and if B⊂Y => {n∈X: f(n)∈B}

f^-1 (A) \ f^−1 (B) = {m∈X: f(m)∈A} \ {n∈X: f(n)∈B} = {l∈X: f(l)∈A and f(l)∉B} => f(l)∈A\B

After here, I could not find a good way to explicitly show my proof. I am having some troubles with scientific mathematical notation. Could you please help me?

Regards

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1 Answer

up vote 3 down vote accepted

I think that you are just not constructing the proof correctly.

Indeed a proof of equality can be done by showing two ways inclusion, but your proof of those is unclear.

Let me do the first part, you will do the second:


Let us see that $f^{-1}(A)\setminus f^{-1}(B)\subseteq f^{-1}(A\setminus B)$. Let $x\in f^{-1}(A)\setminus f^{-1}(B)$ be an arbitrary element, we will show that $x\in f^{-1}(A\setminus B)$.

Since we took $x\in f^{-1}(A)\setminus f^{-1}(B)$ we know that $f(x)\in A$ and $f(x)\notin B$. Therefore $f(x)\in A\setminus B$, and therefore $x\in f^{-1}(A\setminus B)$, as wanted.


It is important to be clear more than it is important to write in cool mathematical symbols and formulae. Use the definitions of the symbols you are given, and make sure you understand them well. Then it is important to understand how to construct a proof by element-chasing: take an element here, unwind the definitions, rewind the definitions and show it is also an element there.

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you are right: "It is important to be clear more than it is important to write in cool mathematical symbols and formulae" Thank you so much! –  Amadeus Bachmann Oct 6 '12 at 11:41
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