# Collection $\cal{C}$ of uncountable subsets of $\mathbb{R}$ such that every countable subset is contained in exactly one member of $\cal{C}$

If $$\kappa$$ is a cardinal and $$X$$ is a set, let $$[X]^\kappa$$ denote the collection of subsets of $$X$$ that have cardinality $$\kappa$$.

Let $$\beta>\omega$$ and $$\beta \leq 2^{\omega}$$. Is there $${\cal C}\subseteq [\mathbb{R}]^\beta$$ such that every member of $$[\mathbb{R}]^\omega$$ is contained in exactly one member of $${\cal C}$$?

• Let B and D be distinct infinite subsets. Then B union D must live in the same member of C as both B and D. So any superset of B must be in the same member as B. This includes the whole space, when it fits. (As Nik Weaver observed in a now deleted comment.) Gerhard "Infinity Is A Strange Place" Paseman, 2019.09.05. – Gerhard Paseman Sep 6 '19 at 5:14

Suppose that every countably infinite subset of $$\mathbb R$$ is contained in exactly one member of $$\mathcal C$$, where $$\mathcal C\subseteq\mathcal P(\mathbb R)$$ and $$\mathbb R\notin\mathcal C$$. Let $$A$$ be a countably infinite subset of $$\mathbb R$$. Choose a set $$S\in\mathcal C$$ such that $$A\subseteq S$$, and choose an element $$t\in\mathbb R\setminus S$$. Consider the countably infinite set $$B=A\cup\{t\}$$. Either $$B$$ is contained in no member of $$\mathcal C$$, or else $$A$$ is contained in two different members of $$\mathcal C$$; either way we have a contradiction.
If every set in $$[\mathbb{R}]^{\aleph_0}$$ is contained in a unique member of $$\mathcal{F}$$, then by induction on $$\aleph_0 \leq \kappa \leq \mathfrak{c}$$, it is easy to see that every set in $$[\mathbb{R}]^{\kappa}$$ is contained in a unique member of $$\mathcal{F}$$. It follows that $$\mathbb{R} \in \mathcal{F}$$ and hence $$\mathcal{F} = \{\mathbb{R}\}$$.
• Nice argument, msybe a little too terse. Let's see. Suppose $\kappa\gt\omega$, $A\in[\mathbb R]^\kappa$, and every infinite subset of $\mathbb R$ of size $\lt\kappa$ is contained in a unique member of $\mathcal C$. Write $A=\bigcup_{\alpha\lt\kappa}A_\alpha$ where $\alpha\lt\beta\implies A_\alpha\lt A_\beta$ and $\aleph_0\le|A_\alpha|\lt\kappa$. Then each $A_\alpha$ is contained in a unique $B_\alpha\in\mathcal C$, so the $B_\alpha$ are all the same, so $A$ is contained in a unique member of $\mathcal C$. Right? – bof Sep 6 '19 at 21:08