How to prove that $\bigcup_{\nu<\theta}\alpha_\nu=\bigcup_{\nu<\theta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]$? Perhaps using transfinite induction I managed to prove that $$\bigcup_{\nu<\theta}\alpha_\nu=\bigcup_{\nu<\theta} \left[ \alpha_\nu\setminus \left( \bigcup_{\xi<\nu} \alpha_\xi \right) \right].$$
If $\theta=0$ the equlity is banally verified.
If $\theta=(\overline\eta+1)$ we suppose that for every $\eta<\theta$ it is $\bigcup_{\nu<\eta}\alpha_\nu=\bigcup_{\nu<\eta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]$ and we observe that $\bigcup_{\nu<\theta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]=\bigcup_{\nu<(\overline\eta+1)}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]=\bigcup_{\nu<\overline\eta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]\cup[\alpha_\overline\eta\setminus(\bigcup_{\xi<\overline\eta}\alpha_\xi)]=[\bigcup_{\nu<\overline\eta}\alpha_\nu]\cup[\alpha_\overline\eta\setminus(\bigcup_{\xi<\overline\eta}\alpha_\xi)]=[\bigcup_{\nu<\overline\eta}\alpha_\nu]\cup\alpha_\overline\eta=\bigcup_{\nu<(\overline\eta+1)}\alpha_\nu=\bigcup_{\nu<\theta}\alpha_\nu$
If $\theta=\bigcup_{\nu<\theta}\nu$ we suppose that for every $\eta<\theta$ it is $\bigcup_{\nu<\eta}\alpha_\nu=\bigcup_{\nu<\eta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]$ and we observe that $\bigcup_{\nu<\theta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]=\bigcup_{\nu<(\bigcup_{\eta<\theta}\eta)}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]=\bigcup_{\eta<\theta}[\bigcup_{\nu<\eta}(\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi))]=\bigcup_{\eta<\theta}(\bigcup_{\nu<\eta}\alpha_\nu)=\bigcup_{\nu<\theta}\alpha_\nu$
So for transfinite induction it is $\bigcup_{\nu<\theta}\alpha_\nu=\bigcup_{\nu<\theta}[\alpha_\nu\setminus(\bigcup_{\xi<\nu}\alpha_\xi)]$.
Is the reasoning correct? If not, could someone prove the equality?
 A: I don't know if your proof is correct, but you're doing way too much work, there's no need for transfinite recursion.
First of all let $A_\nu=\alpha_\nu\setminus\bigcup_{\xi<\nu}\alpha_\xi$. We want to show that $\bigcup_{\nu<\theta}\alpha_\nu=\bigcup_{\nu<\theta}A_\nu$.
By definition we have $A_\nu\subseteq\alpha_\nu$ for every $\nu$, so we also have $\bigcup_{\nu<\theta}A_\nu\subseteq\bigcup_{\nu<\theta}\alpha_\nu$.
To show the reverse inclusion let $x\in\bigcup_{\nu<\theta}\alpha_\nu$, we want to show $x\in\bigcup_{\nu<\theta}A_\nu$. Since $x$ is in the union of the $\alpha_\nu$ it must be in one of them, so we can pick the minimal ordinal $\xi$ such that $x\in\alpha_\xi$. I claim that $x\in A_\xi$, indeed, by minimality of $\xi$ we have that $x\not\in \alpha_\eta$ for $\eta<\xi$, so $x\not\in\bigcup_{\eta<\xi}\alpha_\eta$, and so $x\in\alpha_\xi\setminus\bigcup_{\eta<\xi}\alpha_\eta=A_\xi$. In particular we have $x\in\bigcup_{\nu<\theta}A_\nu$ and so, since $x$ was arbitrary, $\bigcup_{\nu<\theta}\alpha_\nu\subseteq\bigcup_{\nu<\theta}A_\nu$.
Since we proved a double inclusion we have an equality between the two unions.
I suggest you draw some blobs for $\alpha_\nu$ and $A_\nu$, maybe for $\theta$ finite or $\theta=\omega$, to see what's happening, because it's a very intuitive idea that's getting lost in technical details.
