Covering of uncountable cardinal 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 $\alpha = \bigcup y$?
 A: 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:


*

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

*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.
A: 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 $\bigcup x=\omega_1$.  
