Is an arbitrary union of a chain of countable sets countable?

Question

Let $C$ be a chain of countable sets, i.e. $\forall S, T \in C: S \subseteq T \lor T \subseteq S$, and that every $S \in C$ is countable.

Then, is $\bigcup C := \{ t \mid \exists S \in C: t \in S\}$ countable?

The answer is no, and a counter-example is $C = \omega_1$, the first uncountable cardinal/ordinal.

Is there a more elementary example that doesn't involve, say, cardinals and ordinals?

Answer 1

If you wish to assume choice, then you have no way around $\omega_1$. The reason being that any chain has a well-ordered cofinal subset, and thus if all the members of the chain are countable that subset has to be countable or of type $\omega_1$. In the former case the union of the chain is countable, and in the latter it will be of size $\aleph_1$.

But what happens without assuming the axiom of choice? Well. It is consistent that $\Bbb R$ is a countable union of countable sets, or that you have a Russell set, that is a countable union of pairwise disjoint pairs that admit no choice function.

In any such case, you have $\{A_n\mid n<\omega\}$ as a countable family of countable sets (in the Russell set case, finite), then defining $B_n=\bigcup_{k<n}A_n$ gives you that $\{B_n\mid n<\omega\}$ is a countable chain of countable sets whose union is uncountable.

Answer 2

Here is a solution working with the set $\mathbb R$.

First obtain a well-order $<$ for $\mathbb R$.

Consider the set $X = \{ x |\hspace{0.1cm} (-\infty,x] \text{ is not countable }\}$, where the intervals are taken in the well-order.

If $X$ is empty we can let our chain be $\{ (-\infty,x] \hspace{0.1cm}| x \in \mathbb R)$.

If $X$ is not empty let $x$ be the minimum element.

We will let our chain be $\{ (-\infty,y]\hspace{0.1cm} | y<x )$.

every element of the chain is countable, but the union is not, as every element of $(-\infty,x]$ except for $x$ is in the union, and $(-\infty,x]$ is not.