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Let $f:X\rightarrow Y$ where $X$ and $Y$ are sets. Prove that if {$S_\alpha$}$_{\alpha\in I}$ is a collection of subsets of $Y$, then $f^{-1}(\cup_{\alpha\in I}S_\alpha)=\cup_{\alpha\in I}f^{-1}(S_\alpha)$.

Let $x\in f^{-1}(\cup_{\alpha\in I}S_\alpha)$. Then, by definition of preimage, this implies $f(x)\in \cup_{\alpha\in I}S_\alpha$. Which implies that $f(x)$ is in the some subset(s) of $Y$, $S_\alpha$ (where $S_\alpha\subseteq\cup_{\alpha\in I}S_\alpha$), i.e. $f(x)\in S_\alpha$. So, then $x\in f^{-1}(S_\alpha)$ which implies $x\in \cup_{\alpha\in I}f^{-1}(S_\alpha)$.

I still have to show the proof in the other direction, but is this one correct?

Thank you.

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    @BrianM.Scott I will try to clean it up and transition better, thanks!2012-12-09

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Let $x\in f^{-1}(\bigcup_{\alpha\in I}S_\alpha$. Then $f(x)\in \bigcup_{\alpha\in I}S_\alpha$, and in particular, $f(x)\in S_{\alpha^*}$ for at least one $\alpha^*\in I$. Thus $x\in f^{-1}(S_{\alpha^*})$, hence $x\in \bigcup_{\alpha\in I}f^{-1}(S_\alpha)$.

The other direction is similar: Let $x\in \bigcup_{\alpha\in I}f^{-1}(S_{\alpha})$. Then $f(x)\in S_\alpha$ for some $\alpha$, hence $f(x)\in\bigcup_{\alpha\in I}S_\alpha$. This means $x\in f^{-1}(\bigcup_{\alpha\in I}S_\alpha)$.