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I've been independently reading Kunen's newest set theory book for a self-study course. I'm looking at his chapter on cardinal arithmetic, thoroughly reading proofs and working on exercises. After an exercise of the Milner-Rado Paradox, he mentions that types are bounded when it comes to finite unions and states the following exercise:

Let $\kappa$ be an infinite cardinal and $\alpha = \bigcup_{n<c}X_n$, where $c < w$ and type$(X_n) < \kappa^{\omega}$, then $\alpha < \kappa^{\omega}$.

Intuitively, it seems to make sense, but I'm having trouble coming up with a formal proof. Would I use induction on $n$? Any help or hints would be much appreciated. Thanks!

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up vote 2 down vote accepted

You have to prove that if $\operatorname{type}(X_i)<\kappa^\omega$ for $i=1,2$, then $\operatorname{type}(X_1\cup X_2)<\kappa^\omega.$

Visualize $X_1$ as an ordinal $\alpha<\kappa^\omega$. When you add $X_2$ to $X_1$, for each $\beta<\alpha$ you're adding between $\beta$ and $\beta+1$ a set of type at most $\operatorname{type}(X_2)$, and as you're doing this $\operatorname{type}(X_1)$ times, you get that $\operatorname{type}(X_1\cup X_2)\leq\operatorname{type}(X_2)\cdot\operatorname{type}(X_1)<\kappa^\omega.$

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You may enjoy reading the beginning of this answer, where I mention a result of Toulmin characterizing precisely what the possible values of $\mathrm{type}(X_1\cup X_2)$ can be in terms of $\mathrm{type}(X_1)$ and $\mathrm{type}(X_2)$. – Andrés E. Caicedo Sep 27 '13 at 0:35
@CamiloArosemena: Your reasoning makes sense. Once we know that the type of $X_1 \cup X_2$ is less than $\kappa^{\omega}$, then does it automatically follow that $\alpha < \kappa^{\omega}$ since we just keep adjoining an extra $X_i$ to the union? For instance, type$((X_1 \cup X_2) \cup X_3)$ would be less than $\kappa^{\omega}$, and so on. – josh Sep 27 '13 at 0:49
@josh yes, indeed, as type$((X_1\cup X_2)\cup X_3)\leq\operatorname{type}(X_1\cup X_2)\cdot\operatorname{type}(X_3)<\kappa^\omega,$ as we already know that $\operatorname{type}(X_1\cup X_2)<\kappa^\omega$. – Camilo Arosemena Sep 27 '13 at 0:54
@josh you also have to prove that whenever $\alpha,\beta<\kappa^\omega$, then $\alpha\cdot\beta<\kappa^\omega;$ using that $\kappa^\omega=\sup\{\kappa^n:n<\omega\}$ – Camilo Arosemena Sep 27 '13 at 0:55
$\alpha,\beta<\kappa^m$ for large enough $m$, then $\alpha\cdot\beta\leq\kappa^{2m}<\kappa^\omega$ – Camilo Arosemena Sep 28 '13 at 23:10

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