Proof of a theorem for a function of sets

[lookingforhelp]

Let A and B be sets, f: A->B, B1, B2 are subsets of B,
Prove f^-1[B1-B2] = f^-1[B1] - f^-1[B2]
where f^-1[ ] is the inverse image.
I think this will be a two part proof since it's equality, thank you for the help.

 

[stapel]

I think this will be a two part proof since it's equality, thank you for the help.

Yes. Set equality is proven by showing double set inclusion; that is, by showing that the one set is a subset of the other, and then that the other set is a subset of the one.

Let A and B be sets, f: A->B, B1, B2 are subsets of B,
Prove f^-1[B1-B2] = f^-1[B1] - f^-1[B2] where f^-1[ ] is the inverse image.

Pick an element y in B₁ - B₂. By definition of set complements, then y must be in B₁ and not in B₂...

...and so forth.

 

[pka]

Let A and B be sets, f: A->B, B1, B2 are subsets of B,
Prove f^-1[B1-B2] = f^-1[B1] - f^-1[B2]
where f^-1[ ] is the inverse image.

You might see if your classnotes have done these two.
If \(\displaystyle C \subseteq B~\&~D \subseteq B\) then
1) \(\displaystyle {f^{ - 1}}\left( {C \cap D} \right) = {f^{ - 1}}\left( C \right) \cap {f^{ - 1}}\left( D \right)\), 2) \(\displaystyle {f^{ - 1}}\left( {{D^c}} \right) = {\left( {{f^{ - 1}}\left( D \right)} \right)^c}\).

If you have done those the proof is easy.