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Let $\mathcal{F}$ denote a family of countable subsets of $\mathbb{R}$, such that for each $U, V\in\mathcal{F}$ we have that $U\subseteq V$, or $V\subseteq U$. Let $(\mathcal{F}, \preceq)$ denote the inclusion partial order of $\mathcal{F}$.

  1. Is it true that $(\mathcal{F}, \preceq)$ is isomorphic to $(S, \leq)$ where $S$ is a subset of $\mathbb{R}$?

  2. Is it true that $\bigcup_{U\in\mathcal{F}}U$ is a countable set?

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    $\begingroup$ How about Dedekind cuts in $\mathbb Q$? $\endgroup$ Sep 30 at 13:39
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    $\begingroup$ Still no. Fix a well order $\prec$ of type $\omega_1$ on a subset of $\mathbb R$, and let $\mathcal F$ consist of proper initial segments of $\prec$. $\endgroup$ Sep 30 at 13:57
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    $\begingroup$ Emil's second comment also answers question 2. $\endgroup$ Sep 30 at 14:00
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    $\begingroup$ The questions are equivalent, anyway: $\bigcup\mathcal F$ is countable iff $(\mathcal F,\subseteq)$ embeds in $(\mathbb R,\le)$ iff $(\mathcal F,\subseteq)$ has countable cofinality. $\endgroup$ Sep 30 at 14:03
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    $\begingroup$ @EmilJeřábek why not post an answer explicating the argument of your comment? $\endgroup$ Sep 30 at 15:39

1 Answer 1

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$\let\sset\subseteq\def\cF{\mathcal F}\def\R{\mathbb R}\def\Q{\mathbb Q}$The answer to both questions is negative: let $\preceq$ be a well order of type $\omega_1$ on a subset of $\R$, and let $\cF$ consist of all proper initial segments of $\preceq$. Then $\cF$ is a family of countable sets totally ordered by $\subseteq$, but $\bigcup\cF$ is uncountable, and $(\cF,{\sset})$ does not embed in $(\R,{\le})$ as $(\omega_1,{\le})$ does not embed there.

In fact, for any $\cF$ as in the question, the following are equivalent:

  1. $(\cF,{\sset})$ embeds in $(\R,{\le})$.

  2. $\bigcup\cF$ is countable.

  3. $(\cF,{\sset})$ has countable cofinality.

1 → 3: Any subset of $\R$ has a countable cofinal subset.

3 → 2: If $C\sset\cF$ is a cofinal countable subset, then $\bigcup\cF=\bigcup C$ is a countable union of countable sets, hence countable.

2 → 1: Let $I=\bigcup\cF$. Then $\{(x,y):\forall U\in\cF\,(y\in U\to x\in U)\}$ is a total preorder on $I$, hence it includes a total order $\preceq$ on $I$, and every $U\in\cF$ is an initial segment of $\preceq$. Since the lexicographic product $(I,{\preceq})\times2$ embeds in $(\Q,{\le})$, we can find an embedding $f\colon(I,{\preceq})\to(\Q,{\le})$ such that $f(x)>\sup\{f(y):y\prec x\}$ for all $x\in I$. Then $U\mapsto\sup f[U]$ is an embedding of $(\cF,{\sset})$ in $(\R,{\le})$.

Note that the example in the beginning is, in a sense, the worst that can happen: as any total order, $(\cF,{\sset})$ has a well-ordered cofinal subset $C$. If $C$ is not countable, it can only have order type $\omega_1$, as otherwise some element of $\cF$ is uncountable. Thus, $(\cF,{\sset})$ has cofinality $\omega_1$, $|\bigcup\cF|=\aleph_1$, and using a similar argument as above, $(\cF,{\sset})$ embeds in the long line $\omega_1\times[0,1)$.

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