Question from Real Analysis -- making proof cleaner I'm taking my first real analysis class after about maybe 7 years from the last time I took one. I have a question about how to make my proof nicer. The question is: Let $f : X \rightarrow Y$ be a mapping. Let $Y_\alpha,{\alpha \in A}$ be a family of subsets of Y. I need to prove $f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$

So here's how I tried it.

Fix $a \in f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \left\lbrace x \in X : f(x) \in \bigcup_{\alpha \in A} Y_\alpha \right\rbrace$. Then $f(a) \in \bigcup_{\alpha \in A}Y_\alpha$, so there exists some $\beta \in A$ so that $f(a) \in Y_\beta$. This implies that $a \in f^{-1}(Y_\beta)$, and hence $a \in \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$

This shows containment one way. I haven't written it out but I assume to show the other containment is similar. It seems however, that there should be a bit more elegant way to write this than how I did. I think the work I did was right, but is there a way to make it cleaner?
You can write in this way 

$a \in f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \left\lbrace x \in X : f(x) \in \bigcup_{\alpha \in A} Y_\alpha \right\rbrace$ $\iff$ $f(a) \in \bigcup_{\alpha \in A}Y_\alpha$ $\iff$ $ \exists \beta$ s.t $ f(a) \in Y_\beta$   $\iff$ $\exists \beta$ s.t $a \in f^{-1}(Y_{\beta})$$\iff$ $a \in \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$
You can write in this way 

$a \in f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \left\lbrace x \in X : f(x) \in \bigcup_{\alpha \in A} Y_\alpha \right\rbrace$ $\iff$ $f(a) \in \bigcup_{\alpha \in A}Y_\alpha$ $\iff$ $ \exists \beta$ s.t $ f(a) \in Y_\beta$   $\iff$ $\exists \beta$ s.t $a \in f^{-1}(Y_{\beta})$$\iff$ $a \in \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$