Cantor never bores (1)

Given a set of countable sets $K$, such that $K$ is totally ordered by inclusion, videlicet for every $A,B\in K$ either $A\subseteq B$ or $A\supseteq B$. Intuitively, for at every step in this chain one element at least must be added, one expects the set $K$ to be countable as well.

Suppose $K$ is countable. Then the union, $\bigcup K$ is a countable union of countable sets, hence countable. (Suppose $k: \mathbb N \to K$ is an enumeration of $K$ and $f_i: \mathbb N \to k(i)$ enumerations of the elements of the chain. Then $f_0(0), f_1(0), f_0(1), f_2(0), f_1(1), f_0(2), \ldots$ enumerates $\bigcup K$.)

Thus $\bigcup K$ is an upper bound of $K$. In the poset of countable subsets of some set $U$, of which $\bigcup K$ is a subset, every non-empty chain has an upper bound. Hence, using Zorn’s lemma there is a maximal element, say $M$.

Suppose $U$ is uncountable, then there exists a $\star \in U \backslash M$. $M \cup \{\star\}$ is most definitely also countable and $M \subset M \cup \{\star\}$ which contradicts $M$’s maximality. We are forced to conclude that there exists an uncountable chain of countable sets.

Cantor’s set theory keeps surprising.

Update: an example of such a chain is the set of the countable ordinals.

Another update: a “more concrete” example are the downsets in $\mathbb Q$ without the empty set and $\mathbb Q$ itself. These downsets correspond to real numbers, see Dedekind Cuts.