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Let $\kappa$ be an infinite cardinal. Suppose $E \subseteq {\cal P}(\kappa)$ has the following property: for $e_1\neq e_2\in E$ we have $|e_1\cap e_2|= 1$, and suppose $|E| = \kappa$.

Does this imply that at least one of the following statements is true?

  1. there is $e\in E$ with $|e|=\kappa$;
  2. there is $\alpha \in \kappa$ such that $|\{e\in E: \alpha \in e\}| = \kappa$.
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  • $\begingroup$ What is $P(\kappa)\ ?$ $\endgroup$ Commented Nov 21, 2016 at 7:17
  • $\begingroup$ Is the first $\kappa$ of condition 2. simply an arbitrary set of cardinality $\kappa$? $\endgroup$ Commented Nov 21, 2016 at 7:23
  • $\begingroup$ @WłodzimierzHolsztyński 1) the power set of $\kappa$; 2) the operator $|\cdot|$ denotes taking the cardinality. $\endgroup$ Commented Nov 21, 2016 at 13:06

1 Answer 1

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Consider two sets $e,f\in E$, assume that $\max(|f|,|e|)=:\mu<\kappa$, $\{x\}:=e\cap f$. Take arbitrary element $y\in f\setminus x$, it is contained in at most $\mu$ sets from $E$. Indeed, they all have a common element with $e$, and all those common elements are different. So, the elements of $f\setminus x$ are contained totally in at most $\mu\times \mu<\kappa$ sets, thus $x$ is contained in $\kappa$ sets.

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  • $\begingroup$ The size of $e$, not of $f$ is a bound. Because $w\mapsto w\cap e$ is injection from the sets $w$ containing $y$ to $e$. $\endgroup$ Commented Nov 20, 2016 at 16:26
  • $\begingroup$ Doesn't this show something much stronger from the assumptions, like if all sets are smaller than $\kappa$, then all of them intersect in the same element? $\endgroup$
    – domotorp
    Commented Nov 20, 2016 at 19:09
  • $\begingroup$ @domotorp this is an a priori consequence of the OP: if $x$ belongs to $\kappa$ sets, then any set of cardinality less than $\kappa$ contains $x$. $\endgroup$ Commented Nov 20, 2016 at 19:12
  • $\begingroup$ It must be so, but why? $\endgroup$
    – domotorp
    Commented Nov 20, 2016 at 19:22
  • $\begingroup$ Else it has not enough elements to intersect all the sets containing $x$. $\endgroup$ Commented Nov 20, 2016 at 19:24

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