Question about a family of nested countable subsets of $\mathbb{R}$

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?

• How about Dedekind cuts in $\mathbb Q$? Sep 30, 2022 at 13:39
• 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 $\prec$. Sep 30, 2022 at 13:57
• Emil's second comment also answers question 2. Sep 30, 2022 at 14:00
• 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. Sep 30, 2022 at 14:03
• @EmilJeřábek why not post an answer explicating the argument of your comment? Sep 30, 2022 at 15:39

$$\let\sset\subseteq\def\cF{\mathcal F}\def\R{\mathbb R}\def\Q{\mathbb Q}$$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 $$\omega_1\times[0,1)$$.