1. ## Inverse of sets

I have attached a question that I am having trouble with. Thanks to anyone who can help.DOC1.DOC

2. ## Sets and Inverse functions

Hello Mel

Here is the problem.

Suppose $X$ and $Y$ are nonempty sets and $f: X \rightarrow Y$ is a function. Let $A$ and $B$ be subsets of $Y$. Prove that $f^{-1} (A \cup B) = f^{-1}(A) \cup f^{-1}(B)$

We do this by showing that $f^{-1} (A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$ and $f^{-1}(A) \cup f^{-1}(B) \subseteq f^{-1} (A \cup B)$

So, for the first part:

Suppose $x \in f^{-1}(A \cup B)$. Then for some $y \in Y$, $f(x) = y$, and $y \in A \cup B$.

$\Rightarrow y \in A$ or $y \in B$ [Note: throughout this proof 'or' means 'inclusive or'. So $y \in A$ or $y \in B$ means $y \in A$ or $y \in B$ or both.

$\Rightarrow f(x) \in A$ or $f(x) \in B$

$\Rightarrow x \in f^{-1}(A)$ or $x \in f^{-1}(B)$

$\Rightarrow x \in f^{-1}(A) \cup f^{-1}(B)$

So $x \in f^{-1}(A \cup B) \Rightarrow x \in f^{-1}(A) \cup f^{-1}(B)$

$\Rightarrow f^{-1}(A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$ (1)

For the second part:

Suppose $x \in f^{-1}(A) \cup f^{-1}(B)$

Then $x \in f^{-1}(A)$ or $x \in f^{-1}(B)$

Then for some $y \in Y$, $f(x) = y$ and $y \in A$ or $y \in B$

$\Rightarrow y \in A \cup B$

$\Rightarrow x \in f^{-1}(A \cup B)$

So $x \in f^{-1}(A) \cup f^{-1}(B) \Rightarrow x \in f^{-1}(A \cup B)$

$\Rightarrow f^{-1}(A) \cup f^{-1}(B) \subseteq f^{-1} (A \cup B)$ (2)

So, from (1) and (2): $f^{-1} (A \cup B) = f^{-1}(A) \cup f^{-1}(B)$