Cantor never bores (1)

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

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

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

Suppose [tex]U[/tex] is uncountable, then there exists a [tex]\star \in U \backslash M[/tex]. [tex]M \cup \{\star\}[/tex] is most definitely also countable and [tex]M \subset M \cup \{\star\}[/tex] which contradicts [tex]M[/tex]’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 [tex]\mathbb Q[/tex] without the empty set and [tex]\mathbb Q[/tex] itself. These downsets correspond to real numbers, see Dedekind Cuts.

Leave a Reply

Your email address will not be published. Required fields are marked *