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My question is about the following:

Using the Axiom of Choice show that:

If $\kappa\ge\omega$ is a regular cardinal, $\gamma\le\kappa$, and $\langle A_\alpha\mid\alpha\lt\gamma\rangle$ is a sequence of sets each of cardinality less than $\kappa$, then $|\bigcup_{\alpha \lt \gamma} A_\alpha| < \kappa$

I found an answer here: Regular cardinals and unions but I can't see where it uses the axiom of choice. Any help would be much appreciated!

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I think you are choosing $\alpha_{\xi}$ from the set $S_{\xi}=\{\alpha<\kappa: A_{\xi}\subset\alpha\}$ for each $\xi$, which requires the axiom of choice. –  Karl Kronenfeld Apr 1 '13 at 12:55
    
@user1: You never need the axiom of choice to choose from nonempty sets of ordinals. Pick the minimal. –  Asaf Karagila Apr 1 '13 at 13:07
    
@AsafKaragila Obviously, my understanding of AC is weak. But I think I now see why it is not needed here. We can define a function $m:\cal P(\kappa)\setminus\{\emptyset\}\to\kappa$ by selecting the minimal element. For any family $\cal S$ of nonempty subsets of $\kappa$, we can construct a choice function easily, namely the restriction of $m$ to $\cal S$. Thank you. –  Karl Kronenfeld Apr 1 '13 at 13:13
    
@user1: Yes. You will need the axiom of choice if you want to choose from $\mathcal{P(P(\kappa))}\setminus\{\varnothing\}$, though. –  Asaf Karagila Apr 1 '13 at 13:16
    
@AsafKaragila It is interesting to me that one cannot just adapt the lexicographic order to arrange the subsets of $\kappa$. –  Karl Kronenfeld Apr 1 '13 at 13:26
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1 Answer 1

First of all, the definition of regular cardinals does not depend on the axiom of choice, even in $\sf ZF$ regular cardinals are those which cannot be expressed as a "small union of small sets". The proof that every successor cardinal is regular, however, does use the axiom of choice, i.e. there are models of $\sf ZF+\lnot AC$ in which $\aleph_1$ is singular, despite being the successor of $\aleph_0$.

Secondly, you are using the axiom of choice when you choose a well-ordering for $A_\alpha$. The proof would usually go as follows:

Assume without loss of generality that $A_\alpha$'s are pairwise disjoint. Let $\beta_\alpha$ be an order type of a well-order of $A_\alpha$, then we can embed $\bigcup A_\alpha$ into $[0,\beta_0)\cup\bigcup[\beta_\alpha,\beta_{\alpha+1})=\delta$, where $\delta$ is some ordinal. By the regularity of $\kappa$ we have to have $\delta<\kappa$ and therefore $\bigcup A_\alpha$ has size $<\kappa$.

But we cannot always choose such sequence of well-orders. For example, the union of countably many sets of size $2$ might not be well-orderable at all. Or the countable union of countable sets might have cardinality $\aleph_1$, in which case $\aleph_1$ is not regular.

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IIRC things go pretty wild with respect to the alephs even in the 'well-behaved' choiceless models, right? e.g. doesn't AD imply that $\text{cof}(\omega_n) = \omega_2$ for all $n\gt 2$? –  Steven Stadnicki Apr 1 '13 at 23:49
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@Steven: I wouldn't dream calling any model of $\sf ZF+AD$ "well-behaved" (in terms of choice, anyway). But yes, there is quite a gap between $\omega_2$ and the next regular cardinal in these models. Do note, however, that we can make $\aleph_1$ singular without any large cardinals. If we want two successive singular cardinals then we already need some very large cardinals in the background. –  Asaf Karagila Apr 1 '13 at 23:52
    
Thanks very much for your help! I have learnt that $\kappa$ is a regular cardinal if cf($\kappa)=\kappa$ so are you saying this is equivalent to saying that $\kappa$ cannot be expressed as a "small union of small sets"? –  ELT Apr 3 '13 at 20:26
    
@ELT: There are two ways to define cofinality, one is by minima order type of an unbounded set; the other by the least cardinality of an index set for which small sets give $\kappa$. For ordinals these are indeed equivalent. The proof is essntiallly in my answer. As for the axiom of choice, it is only needed to show that successors are regular. –  Asaf Karagila Apr 3 '13 at 20:50
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