# Some set proofs

1. Jan 21, 2007

### Geekster

Disclaimer: I might have some problems getting my LaTeX code to work properly so please bear with me while I figure out how to properly use the forum software.

1. The problem statement, all variables and given/known data
The exercise is to prove the following statements.

Suppose that $f:X \rightarrow Y$, the following statement is true.
If $\{G_{\alpha} : \alpha \in A\}$ is an indexed family of subsets of Y, then $f^{-1}(\bigcup_{\alpha \in A} G_\alpha) =\bigcup_{\alpha \in A} f^{-1}( G_\alpha)$.

2. Relevant equations

The relevant information in this case is the definitions. In this case I need to know what the definition of an inverse function is.

DEF: Suppose $f:X \rightarrow Y$ and $A \subset Y$. $f^{-1}(A) = \{x \in X: f(x) \in A\}$

3. The attempt at a solution

The solution I've been looking thus far is a point wise argument.

Choose $t \in f^{-1}(\bigcup_{\alpha \in A} G_\alpha)$. So by definition we know
$t \in \{x \in X: f(x) \in (\bigcup_{\alpha \in A} G_\alpha)\}$. And here is where I'm kind of stuck. I need to some how get my chosen element to be in the other set, therefore making the one set a subset of the other. Then complete the converse of the argument to finish the proof.

Any ideas on where I should be looking, or what I should be thinking about here?

Last edited: Jan 21, 2007
2. Jan 21, 2007

### StatusX

$f(x) \in \cup_{\alpha \in A} G_\alpha$ is equivalent to "there is some $\alpha \in A$ such that $f(x) \in G_\alpha$". Also, you don't have to worry about doing the converse if every step you take is an equivalence.