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Let $\kappa<\lambda$ be cardinals where $\lambda \geq \aleph_0$ and $\kappa>1$. Is there necessarily a set ${\cal S}\subseteq {\cal P}(\lambda)$ with the following properties?

  1. ${|\cal S}| > 1$,
  2. $|A| = \lambda$ for all $A\in {\cal S}$, and
  3. $B\subseteq \lambda$ with $|B| = \kappa$ implies $\big|\{A \in {\cal S}: B \subseteq A\}\big| = 1$.
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    $\begingroup$ Should "$A\subseteq B$" be "$A\supseteq B$" in your last bulletpoint? $\endgroup$
    – Noah Schweber
    Commented Jun 12, 2023 at 19:49
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    $\begingroup$ Suppose $\kappa$ is infinite. Then let $L:P_\kappa(\lambda)\rightarrow S$ be the unique function with $B\subseteq L(B)$ for $B\in P_\kappa(\lambda)$. Then if $A\subseteq B$, then since $A\subseteq B\subseteq L(B)$ we know that $L(A)=L(B)$ whenever $A\subseteq B$. In particular, if $A,B\in P_\kappa(\lambda)$, then $L(A)=L(A\cup B)=L(B)$ which contradicts (1). There is a reason we don't hear about infinite Steiner systems. The case where $\kappa$ is finite should follow from the Löwenheim–Skolem theorem. $\endgroup$
    – Joseph Van Name
    Commented Jun 12, 2023 at 20:13
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    $\begingroup$ To make the question more interesting and non-trivial yet still a generalization of the finite case, instead of requiring for $|A|=\lambda$ for $A\in S$, we can instead require that that if $B$ has order type $\kappa$ instead of simply cardinality $\kappa$. And here, we can let $\kappa$ be just an ordinal instead of a cardinal (I did not think about this too much, so maybe it has a simple solution in this case too). $\endgroup$
    – Joseph Van Name
    Commented Jun 12, 2023 at 20:34
  • $\begingroup$ I think the case where we are working with just the order type we won't have any Steiner systems for infinite $\kappa$. $\endgroup$
    – Joseph Van Name
    Commented Jun 12, 2023 at 20:46

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