Proving f^-1(Y ∩ Z) = f^-1(Y) ∩ f^-1(Z)

[cateater2000]

Hi

let f be a function from set A into X, and Y,Z c X. Prove the following

f^-1(YandZ)=f^-1(Y)andf^-1(Z);

any tips would be great

 

[rgrig]

When you need to prove that two sets are equal, $$A = B$$, as in your problem the simplest trick you can use is to show that any $$a \in A$$ is also an element of B and viceversa.

 

[cateater2000]

I'm really confused with how to do it with function inverses, thanks for the help i'll try to figure it out

 

[phoenixthoth]

Hi

let f be a function from set A into X, and Y,Z c X. Prove the following

f^-1(YandZ)=f^-1(Y)andf^-1(Z);

any tips would be great

I think you mean that f is a one-to one function from A to X. Is that what you mean by "into"? Also, I think you mean that Y and Z are both subsets of X. Stop me if I'm wrong.

Let x∈f^-1(Y ^ Z). We wish to show that x∈f^-1(Y)&f^-1(Z). Then once we do that, we wish to start by letting x be in f^-1(Y)&f^-1(Z) and show that that implies x∈f^-1(Y ^ Z).

This is what the last poster was writing about.

Now the thing to remember is that x∈f^-1(U) if and only if f(x)∈U.

 

[mathwonk]

I think he does not mean the function is one to one, but "and" seems to mean intersection. and f^(-1) just means preimage.


then this is a corolalary of the usual tautological fact that pullback or inverse image of sets is a boolean homomorphism, i.e. preserves both intersections and unions.

 

[cateater2000]

Hey thanks for youe help I got it now. It's actually really easy thankyou very much :)