TITLE: How to prove the Milner-Rado Paradox?
QUESTION [11 upvotes]: For every ordinal $\alpha<\kappa^+$ there are sets $X_n\subset\alpha$ $(n\in\Bbb{N})$ such that $\alpha=\bigcup_n X_n$ and for each $n$ the order-type of $X_n$ is $\le\kappa^n$.
[By induction on $\alpha$, choosing a sequence cofinal in $\alpha$.]

I tried to prove this problem by using transfinite induction.

For $\alpha=0$, this statement is trivial. 
If $\alpha$ has sets $X_n\subset \alpha$ satisfies $\alpha=\bigcup_n X_n$, $\text{order-type of }X_n\le\kappa^n$, we define
  $X'_0=\{\alpha\}$, $X_{n+1}'=X_n$. Then $\{X_n'\}$ satisfies $\bigcup_n X_n'=\alpha+1$, $\text{order-type of }X'_n \le \kappa^n$.

But I don't know how to proceed the proof. Thanks for any help.

REPLY [4 votes]: $\newcommand{\cf}{\operatorname{cf}}$There’s no problem if $\alpha\le\kappa$, so suppose that $\alpha$ is a limit ordinal such that $\kappa<\alpha<\kappa^+$. Let $\lambda=\cf\alpha$, and let $\langle\alpha_\xi:\xi<\lambda\rangle$ be an increasing sequence cofinal in $\alpha$. For each $\xi<\lambda$ let $\{X_n^\xi:n\in\omega\}$ be a Milnor-Rado decomposition of $\alpha_\xi$, and let $\{X_n^\lambda:n\in\omega\}$ be a decomposition of $\lambda$.
For $n\in\omega$ let
$$X_{2n}=\bigcup_{\eta\in X_n^\lambda}\left(X_n^\eta\cap\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_\xi\right)\;;$$
the order type of each set 
$$X_n^\eta\cap\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_\xi$$
is clearly at most that of $X_n^\eta$, which by hypothesis is at most $\kappa^n$. Moreover, if $\eta<\theta$, 
$$\beta\in X_n^\eta\cap\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_\xi\;,\quad\text{and}\quad\gamma\in X_n^\theta\cap\alpha_\eta\setminus\bigcup_{\xi<\theta}\alpha_\xi\;,$$
then $\beta<\delta$. Or in brief,
$$X_n^\eta\cap\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_\xi<X_n^\theta\cap\alpha_\eta\setminus\bigcup_{\xi<\theta}\alpha_\xi\;.$$
Since the order type of $X_n^\lambda$ is at most $\kappa^n$, that of $X_{2n}$ is at most $\kappa^n\cdot\kappa^n=\kappa^{2n}$. Let $X_{2n+1}=\varnothing$, whose order type is certainly no more than $\kappa^{2n+1}$. It only remains to show that $\alpha=\bigcup_{n\in\omega}X_n$.
If $\beta<\alpha$, let $\eta<\lambda$ be minimal such that $\beta<\alpha_\eta$. Then $\beta\in\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_xi$, and $\alpha_\eta=\bigcup_{n\in\omega}X_n^\eta$, so 
$$\beta\in X_n^\eta\cap\alpha_\eta\setminus\bigcup_{\xi<\eta}\alpha_\xi\subseteq X_{2n}$$
for some $n\in\omega$.