Timeline for Infinite Steiner systems

6 events

when toggle format	what		by	license	comment
Jun 12, 2023 at 20:46	comment	added	Joseph Van Name		I think the case where we are working with just the order type we won't have any Steiner systems for infinite $\kappa$.
Jun 12, 2023 at 20:34	history	edited	Dominic van der Zypen	CC BY-SA 4.0	edited body
Jun 12, 2023 at 20:34	comment	added	Joseph Van Name		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).
Jun 12, 2023 at 20:13	comment	added	Joseph Van Name		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.
Jun 12, 2023 at 19:49	comment	added	Noah Schweber		Should "$A\subseteq B$" be "$A\supseteq B$" in your last bulletpoint?
Jun 12, 2023 at 19:40	history	asked	Dominic van der Zypen	CC BY-SA 4.0