Equality of functions of unions

cope · #1 $Old$ November 5th, 2009, 07:52 AM

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?

Plato · #2 $Old$ November 5th, 2009, 08:13 AM

Part a) is correct.

For part b) be careful! You cannot use the "if and only if" the way you did in a).

cope · #3 $Old$ November 5th, 2009, 10:49 AM

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