Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

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 $\operatorname{type}(X_1\cup X_2)\leq\operatorname{type}(X_2)\cdot\operatorname{type}(X_1)<\kappa^\omega.$

share|improve this answer
1  
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 $\mathrm{type}(X_2)$. –  Andres 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
1  
$\alpha,\beta<\kappa^m$ for large enough $m$, then $\alpha\cdot\beta\leq\kappa^{2m}<\kappa^\omega$ –  Camilo Arosemena Sep 28 '13 at 23:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.