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Let $S_\circ$ be a family of sets in ZFC, indexed by ordinals, with $S_\alpha \subsetneq S_\beta$ for $\alpha < \beta$. Is it possible to have some uncountable $\gamma$ with $S_\gamma$ countable?

Asaf Karagila
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Damian
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  • Hello, what have you tried so far? – B E I R U T Apr 24 '21 at 17:11
  • Is the index set downward closed? – Joe Apr 24 '21 at 17:17
  • $S_\circ$ is defined for all ordinals. You might cut off some convenient initial segment. – Damian Apr 24 '21 at 17:39
  • What are “all ordinals”? Do you mean all ordinals less than some ordinal? If you can “cutoff some convenient initial segment”, then I would cutoff all of the countable ordinals. – Joe Apr 24 '21 at 17:54
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    All ordinals as in the proper class of ordinals. I don't know if the definition of downward closed is applicable to proper classes, so if you really need an index set, you may use an arbitrarily large ordinal. – Damian Apr 24 '21 at 18:08
  • Oh, I have never studied classes, and only know the set theory that I needed for my analysis courses, in which I learned why we cannot have a set of all ordinals. But for your question, the index cannot start at an uncountable ordinal, right? Otherwise it seems like the answer would be trivially yes. – Joe Apr 24 '21 at 21:54
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    @Joe It starts at zero. I think when Damian said "you might cut off some convenient initial segment" they meant cut off and keep, not cut off and throw away. – spaceisdarkgreen Apr 24 '21 at 22:01

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No, there's an injection $f:\gamma\to S_\gamma$. Let $f(\alpha)$ be some element of $S_{\alpha+1}-S_\alpha.$

spaceisdarkgreen
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    +1. Two quick comments for the OP: $(1)$ Note that while this might look like it requires choice, it doesn't: if $S_\gamma$ is countable (or indeed well-orderable at all) we can canonically pick an element of $S_{\alpha+1}\setminus S_\alpha$ using a fixed enumeration of $S_\gamma$. $(2)$ Meanwhile, we can have an uncountable $\subseteq$-chain of countable sets, see e.g. here; the saving grace is that the ordertype of such a chain must have countable cofinality. – Noah Schweber Apr 24 '21 at 17:53
  • @Noah Schweber, that’s blowing my mind. For some reason, I never thought about how for $r \in \mathbb{R}$, $S_r = {q \in \mathbb{Q} | q < r }$ gives an uncountable family of sets such that $r_1<r_2 \implies S_{r_1} \subsetneq S_{r_2}$, and yet they are all countable. The fact that between any two reals there are infinitely many rationals never seemed mind blowing, since choosing two reals involves choosing a positive interval size, and the whole real line could be covered by countably many intervals of that size. But this uncountable sequence of countable sets is quite jarring. – Joe Apr 24 '21 at 21:49