Let $\alpha$ be an uncountable cardinal number and $x$ a countable set of countable subsets of $\alpha$ with $\alpha = \bigcup \bigcup x$. How can be shown that there exists a countable subset $y$ of $\alpha$ such that $\alpha = \bigcup y$?

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    $\begingroup$ Hint: take $y = \cup x$ . $\endgroup$ – Shervin Sorouri Dec 19 '19 at 13:27
  • $\begingroup$ Where do we use that $\alpha$ is a cardinal? And why is $\cup x$ countable? $\endgroup$ – Fblthp Dec 19 '19 at 14:15
  • $\begingroup$ Any countable union of countable sets is countable by the axiom of choice. So $\cup x$ is countable. The fact that $\alpha$ is a cardinal is not used here. $\endgroup$ – Shervin Sorouri Dec 19 '19 at 14:28
  • $\begingroup$ Does this still hold if we consider ZF without the axiom of choice? $\endgroup$ – Fblthp Dec 19 '19 at 14:31
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    $\begingroup$ @Shervin Without choice your suggestion does not work. $\endgroup$ – Andrés E. Caicedo Dec 19 '19 at 15:00

If $x$ is a set of ordinals, then $\bigcup x=\sup x$, and if $x$ is a set of sets of ordinals, then $\bigcup\bigcup x=\sup\{\sup y\mid y\in x\}$. So in other words, you statement can be recast in the following way:

Suppose that there is a countable family of sets $y_n\subseteq\alpha$ such that $\alpha=\sup\left(\bigcup_{n<\omega}y_n\right)$, then there is a countable $y\subseteq\alpha$ such that $\alpha=\sup y$.

There are two options:

  1. There is some $y\in x$ such that $\sup y=\alpha$. Then we are done, since $y$ is countable set by definition.

  2. Otherwise, let $\alpha_n=\sup y_n$ for $y_n\in x$, then $y=\{y_n\mid n\in\omega\}$, or equivalently $y=\{\bigcup z\mid z\in x\}$, is such that $\bigcup y=\alpha$ and $y$ is countable since there is a surjection from $x$ onto $y$. (Note that this holds even without the Axiom of Choice, since a surjection from a countable set is finite or countable even in ZF.)

Finally, note that this is in fact equivalent to $\alpha$ having a countable cofinal set, since in that case taking a countable cofinal subset $y=\{y_n\mid n<\omega\}$ and taking $x=\{\{y_n\}\mid n<\omega\}$ works.


Asaf's is clearly the right way of phrasing things. Anyway, let's clean this up a bit.

Enumerate $x$, say $x=\{x_n\}_n$. If any of its members is cofinal in $\alpha$, you are done - the set you want is any member of $x$ that is cofinal in $\alpha$. Otherwise, let $\alpha_n:=\sup x_n$ and note that $\{\alpha_k:k\in\omega\}$ is cofinal in $\alpha$, since $\bigcup\bigcup x\subseteq\sup_k\alpha_k$.

For instance, (without choice) we could have $\omega_1$ be singular and let $x$ be countable and cofinal in $\omega_1$. In this case we already have $\bigcup x=\omega_1$.


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