# Mapping Proof

• Jan 31st 2008, 09:41 PM
Jhevon
Mapping Proof
For some reason i'm having trouble with this problem...:o

Prove that a mapping $\displaystyle \alpha : S \to T$ is one-to-one iff $\displaystyle \alpha (A \cap B) = \alpha (A) \cap \alpha (B)$ for every pair of subsets $\displaystyle A$ and $\displaystyle B$ of $\displaystyle S$.

we define something like $\displaystyle \alpha (A)$ by $\displaystyle \alpha (A) = \{ \alpha (x) : x \in A \}$

Thanks guys
• Jan 31st 2008, 11:21 PM
a mapping $\displaystyle \alpha : S \to T$ is one-to-one if $\displaystyle \alpha (A \cap B) = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S.
if $\displaystyle \alpha (A \cap B) \not = \alpha (A) \cap \alpha (B)$ for every pair of subsets A and B of S then $\displaystyle \alpha$ is not one to one.
intuitively, $\displaystyle \alpha (A \cap B) \subseteq \alpha (A) \cap \alpha (B)$
So $\displaystyle \alpha (A \cap B) \not = \alpha (A) \cap \alpha (B)$ implies that there is some element in $\displaystyle \alpha (A) \cap \alpha (B)$ that is not in $\displaystyle \alpha (A \cap B)$. So there is an element a in A and an element b in B such that $\displaystyle \alpha (a) = \alpha (b)$ but $\displaystyle a,b \not \in A \cap B$ so $\displaystyle a \not = b$ so $\displaystyle \alpha$ is not one-to-one.
To go the other way, we know that the function is not one-to-one so for some $\displaystyle a \not = b,$$\displaystyle \alpha (a) = \alpha (b)$. using A={a} and B = {b}, $\displaystyle \alpha (A \cap B) = \alpha (\oslash) = \oslash \not = \alpha (A) \cap \alpha (B)$.