0
$\begingroup$

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 ${\cal C}$?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Sep 6 '19 at 5:14
6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mathcal{F} = \{\mathbb{R}\}$.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ – bof Sep 6 '19 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.