Prove the following, $f(\cup_{\alpha} E_{\alpha}) = \cup _{\alpha} f(E _{\alpha})$... In my real analysis class I'm suppose to prove the following...
12)Let $f$ be a function from X into Y.
a)If $ \lbrace E_ \alpha \rbrace _{\alpha \in A}$ is a family of subsets of X, prove that $f(\cup_{\alpha} E_{\alpha}) = \cup _{\alpha} f(E _{\alpha})$.
b)If $ \lbrace B_ \alpha \rbrace _{\alpha \in A}$ is a family of subsets of Y, prove that $f^{-1}(\cap_{\alpha} B_{\alpha}) = \cap _{\alpha} f^{-1}(B _{\alpha})$. 
I've been really struggling to figure out how to prove these.
This is what I have for a)
Let $y \in f(\cup_{\alpha} E_{\alpha})$ then $\exists x \in \cup_{\alpha} E_{\alpha} : f(x) =y$. Thus $x \in \cup_{\alpha} E_{\alpha}$. So $ x \in  E_{\alpha _0}$ for some $\alpha _0$. Then $ y \in f(\alpha_0)$. Thus $y \in f(E_{ \alpha_0}) \subset \cup_{ \alpha} f (E_{ \alpha})$. 
 $_{\square}$
But is this even correct? 
Here is what I have for b) 
Let $x \in f^{-1}(\cap_{\alpha} B_{\alpha})$ then $\exists y \in \cap_{\alpha} B_{\alpha} : f(y) = x$.
That is, $f(x) \in \cap_{\alpha} B_{\alpha}$. Thus $f(x) \in B_{\alpha}$ for all $\alpha$. Therefore $x \in f^{-1}B_{\alpha}$ for all $\alpha$. So $x \in \cap _{\alpha} f^{-1}(B _{\alpha})$. $_{\square}$
But is this even correct?
If someone could please help to tell/explain if I'm doing anything wrong that would be very helpful. 
 A: Your comment on angryavian's answer with reverse inclusions has some issues.  I cannot comment so I will give a full answer:
In part a) you have $y \in f(\alpha_0)$ which is incorrect because $\alpha_0$ is an index and $y = f(x)$ for some $x$.   Just take that out.  For b) replace "$f(y) = x$" with "$f(x) = y$".  As angryavian said.
Be careful not to mix up your notation, you are using $y$ as an image of an element $x$.  You need the reverse inclusions:
a) Want to show: $\bigcup_{\alpha}f(E_\alpha) \subseteq f(\bigcup_{\alpha}E_\alpha)$
Let $ y \in \bigcup_{\alpha}f(E_a\alpha)$.  So $y \in f(E_{\alpha_0})$ for some $\alpha_0$.  So $\exists x \in E_{\alpha_0} : y = f(x)$.  $x \in E_{\alpha_0} \implies x \in \bigcup_{\alpha}E_{\alpha}$.  So $y \in f(\bigcup_{\alpha}E_{\alpha})$. $\square$
b)  Want to show: $f^{-1}(\bigcap_{\alpha}B_{\alpha}) \subseteq \bigcap_{\alpha}f^{-1}(B_{\alpha})$
Let $x \in f^{-1}(\bigcap_{\alpha}B_{\alpha})$. So $\exists y \in \bigcap_{\alpha}B_{\alpha} : f(x) = y$.  Furthermore, $\space y \in B_{\alpha}$ for all $\alpha$.  So $\forall\alpha : x \in f^{-1}(B_{\alpha})$.  Thus $x \in \bigcap_{\alpha}f^{-1}(B_{\alpha})$. $\square$
A: For a), your work is correct after removing the sentence "Then $y \in f(\alpha_0)$." To finish part a) you need to do show the reverse inclusion $f(\bigcup_\alpha E_\alpha) \supseteq \bigcup_\alpha f(E_\alpha)$ in a similar fashion.
For b), you should replace "$f(y)=x$" with $f(x)=y$. Otherwise everything is correct. To finish the proof, you again need to show the reverse inclusion in a similar fashion.
