Let X be a non-empty set, and let S contain all countable subsets of X. Partially order S by inclusion. Let C be a totally ordered subset ("chain") of S, and let(adsbygoogle = window.adsbygoogle || []).push({});

[tex]U = \cup_{E \in C} E[/tex]

It appears that U is not always countable: if it were, U would be an upper bound of the chain C, and U would be in S. From Zorn's lemma [by the way, I'm assuming AC throughout] S would have a maximal countable set M. Letting X be any uncountable set, there would then be some point x in X but not in M. Adding x to M produces a larger but still countable set, contradicting M's maximality.

Now I'm confused, since this is very unintuitive to me. For instance, I can't for the life of me construct an explicit chain C letting X be the real numbers. It's certainly the case that a chain resulting in an uncountable U must then be over uncountably many distinct sets E, which should give some uncountable set "uncountably far up" the chain. Of course, this isn't rigorous while the above is, so it must be wrong. So... could someone un-confuse me, perhaps by providing an explicit example of a set of countable sets totally ordered by inclusion whose union is uncountable? Thanks!

To be extra explicit, C is totally ordered. The analogous result without this condition is trivial: the union of all singleton subsets of the real numbers is uncountable.

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# Uncountable union of a chain of countable sets can be uncountable?

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