# 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 $$\alpha = \bigcup y$$?

• Hint: take $y = \cup x$ . – Shervin Sorouri Dec 19 '19 at 13:27
• Where do we use that $\alpha$ is a cardinal? And why is $\cup x$ countable? – Fblthp Dec 19 '19 at 14:15
• 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. – Shervin Sorouri Dec 19 '19 at 14:28
• Does this still hold if we consider ZF without the axiom of choice? – Fblthp Dec 19 '19 at 14:31
• @Shervin Without choice your suggestion does not work. – 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 $$\bigcup x=\omega_1$$.