Equality of functions of unions

• November 5th 2009, 06:52 AM
cope
Equality of functions of unions
I've been having some trouble with this question, I'm hoping someone could tell me if I'm on the right path.

Given the sets A, B, E, F, M and N where $A,B \subset M$ and $E,F \subset N$, and the function $f : M \rightarrow N$, show whether the following equalities hold:

a) $f^{-1}(E \cup F) = f^{-1}(E) \cup f^{-1}(F)$
b) $f(A \cup B) = f(A) \cup f(B)$

For part a I've done this:

Let $x \in f^{-1}(E \cup F)$
$\Leftrightarrow f(x) \in (E \cup F)$
$\Leftrightarrow f(x) \in E \vee f(x) \in F$
$\Leftrightarrow x \in f^{-1}(E) \vee x \in f^{-1}(F)$
$\Leftrightarrow x \in ( f^{-1}(E) \cup f^{-1}(F) )$

If I'm understanding this right A,B and E,F are proper subsets of M and N respectively, so the function f could map values between M and N that aren't in the union of A,B or E,F. But in part b) for instance f is being applied to a subset of M (A u B) and so any values outside this subset but still part of set M are irrelevant for the proof, right?

So is the proof for a) complete?
• November 5th 2009, 07:13 AM
Plato
Part a) is correct.

For part b) be careful! You cannot use the "if and only if" the way you did in a).
• November 5th 2009, 09:49 AM
cope
Thanks for that. I think I've got b) as well;

Let $y \in f(A \cup B)$, then $\exists x \in A \vee B$ such that $y=f(x)$. Then:

$(y=f(x) : x \in A) \vee (y=f(x) : x \in B)$
$\Leftrightarrow y \in f(A) \vee y \in f(B)$
$\Leftrightarrow y \in f(A) \cup y \in f(B)$

Therefor $f(A \cup B) = f(A) \cup f(B)$