Is there a family $\{A_\alpha:\alpha<2^{\omega_1}\}\subset [\omega_1]^{\omega_1}$ in ZFC such that for each countable set $I\subset 2^{\omega_1}$ and $\alpha\in 2^{\omega_1}\setminus I$ we have $$A_\alpha\not\subset\bigcup\{A_\beta:\beta\in I\}.$$

Of course, we can find such a family of size $\omega_2$ instead of $2^{\omega_1}$.

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    $\begingroup$ So you are looking for a $\sigma$-ideal which is generated by the maximal number of sets possible. Right? $\endgroup$ – Asaf Karagila May 5 '16 at 11:36
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    $\begingroup$ Under CH we can do this, even if $2^{\omega_1}$ is very large, by labeling the nodes of the tree $2^{<\omega_1}$ with ordinals below $\omega_1$, and then taking all paths through the tree. $\endgroup$ – Joel David Hamkins May 5 '16 at 12:03
  • $\begingroup$ What does $[\omega_1]^{\omega_1}$ stand for? $\endgroup$ – YCor May 5 '16 at 12:33
  • $\begingroup$ @YCor: All subsets of $\omega_1$ whose cardinality is exactly $\omega_1$. $\endgroup$ – Asaf Karagila May 5 '16 at 13:28
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    $\begingroup$ If $\{A_{\alpha} : \alpha \in 2^{\omega_1}\} \subseteq [\omega_1]^{\omega_1}$ is almost disjoint modulo countable sets, then it satisfies this property. As Joel noted, such almost disjoint collections exist if CH holds, but also under some other cardinal arithmetic assumptions, e.g. as long as $2^{\omega_1}>2^{\omega}$ and $2^{\omega}<\aleph_{\omega_1}$. See Baumgartner's "Almost-Disjoint Sets, the Dense Set Problem, and the Partition Calculus (1976)" for a reference. However, one can force a model with $2^{\omega}=2^{\omega_1}=\omega_3$ with no such almost disjoint collections. $\endgroup$ – Geoff Galgon May 6 '16 at 6:17

Today I could prove that the existence of such family is not provable from ZFC.

Theorem. If GCH holds in $V$, and $\mu\ge \omega_3$ then in $V^{Fn(\mu,2)}$ the following holds: If $\{A_i:i<{\omega}_3\}\subset [{\omega}_1]^{{\omega}_1}$, then there is $I\in [\omega_3]^\omega$ and $\alpha\in \omega_3\setminus I $ such that $$A_{\alpha}\subset \bigcup_{i\in I}A_i.$$

Proof. An ${Fn(I,2)}$-name $\dot B$ of a subset of $\omega_1$ is called nice if for each ${\nu}<\omega_1$ there is an antichain $B_{\nu}\subset {Fn(I,2)}$ such that $$ \dot B=\{\langle p,\check{{\nu}}\rangle:{\nu}\in {\omega_1}\land p\in B_{\nu}\}= \bigcup\{B_{\nu}\times\{\check{{\nu}}\}:{\nu}\in {\omega_1}\}. $$ We let $supp(\dot B)=\bigcup\{dom(p):p\in\bigcup\limits_{{\nu}<{\omega_1}}B_{\nu}\}$.

It is well-known that every set of ordinals in $V[G]$ has a nice name in $V$.

If ${\varphi}$ is a bijection between two sets $I$ and $J$ then ${\varphi}$ lifts to a natural isomorphism between ${Fn(I,2)}$ and ${Fn(I,2)}$, which will be also denoted by ${\varphi}$, as follows: for $p\in {Fn(I,2)}$ let $dom({\varphi}(p))={\varphi}''dom(p)$ and ${\varphi}(p)({\varphi}({\xi}))=p({\xi})$. Moreover ${\varphi}$ also generates a bijection between the nice ${Fn(I,2)}$-names and the nice $Fn(J,2)$-names: if $\dot A$ is a nice $Fn(I,2)$-name then let ${\varphi}(\dot A )=\{\langle{\varphi}(p),\hat{{\xi}}\>:\langle p,\hat{{\xi}}\rangle \in\dot B\}$.

If $I$ and $J$ are sets of ordinals with the same order type then ${\varphi}_{I,J}$ is the natural order-preserving bijection from $I$ onto $J$.

For each $i<{\omega}_3$ pick a nice $Fn(\mu,2)$-name $\dot A_i$ for $A_i$ in $V$. Then $S_i=supp(A_i)\in [\mu]^{\le \omega_1}$.

Since $2^{\omega_1}=\omega_2$ in $V$ there is a subset $K\in [\omega_3]^{\omega_3}$ such that

(1) $\{S_i:i\in K\}$ forms a $\Delta$-system with kernel $S$,

(2) every element of the family $\{S_i:i\in K\}$ has the same order type

(3) $\varphi_{S_i,S_j}$ is the identity on $S$ for all $i,j\in K$.

(4) $\varphi_{S_i,S_J}(\dot A_i)=\dot A_j$ for $i,j\in K$.

Let $I\in [K]^\omega$ and $\alpha \in K\setminus I$. Assume on the contrary that $p\in Fn(\mu,2)$ and $\zeta<\omega_1$ such that $$ p\Vdash \check \zeta\in \dot A_{\alpha}\setminus \bigcup_{i\in I}\dot A_i. $$ Pick $\beta \in I$ such that $dom(p)\cap (S_\beta\setminus S)=\emptyset.$ Let $$ r=p\cup \varphi_{S_\alpha, S_\beta}(p\restriction S_\alpha). $$ Since $dom(p)\cap (S_\beta\setminus S)=\emptyset$, $r$ is a condition. Moreover, $p\Vdash \check \zeta\in \dot A_{\alpha}$, so $p\restriction S_\alpha \Vdash \check \zeta\in \dot A_{\alpha}$, and so $\varphi_{S_\alpha, S_\beta}(p\restriction S_\alpha )\Vdash \check \zeta\in \dot A_{\beta}$.

Thus $r\Vdash \zeta\in \bigcup_{i\in I}\dot A_i.$ Contradiction.

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    $\begingroup$ I am guessing that a better statement would that the existence of such family is not provable from ZFC. As it stands, the first sentence can be understood that ZFC proves there is no such family. $\endgroup$ – Asaf Karagila May 11 '16 at 23:14
  • $\begingroup$ @AsafKaragila Done. $\endgroup$ – Lajos Soukup May 12 '16 at 13:38
  • $\begingroup$ We were looking for a weak P-space $X\subset 2^{\omega_1}$ of size $2^{\omega_1}$ in ZFC. I asked the question above about families because if there is such a family $\mathcal A$ then we also have a space $X$. However, using the argument of the proof above now I can also prove that the existence of such a space $X$ is not provable in ZFC. $\endgroup$ – Lajos Soukup May 12 '16 at 13:51

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