# Subset of $[\kappa]^{<\kappa}$ with linear intersection

For any cardinal $$\kappa$$, let $$[\kappa]^{<\kappa}$$ denote the collection of subsets of $$\kappa$$ with cardinality $$<\kappa$$. Is there an infinite cardinal $$\kappa$$ and $${\cal C}\subseteq [\kappa]^{<\kappa}$$ with the following properties?

1. $$c \neq d \in {\cal C} \implies |c\cap d|= 1$$,
2. For all $$\alpha\in \kappa$$ we have $$|\{c\in {\cal C}: \alpha\in c\}|>1$$.
• Take a complete graph on $\kappa$, for any $\kappa>2$. – Wojowu Jun 2 '20 at 8:35
• @bof Oh how silly of me, I misread that as an inequality $\leq 1$. – Wojowu Jun 2 '20 at 11:05

There is no such $$\kappa$$. Consider an infinite cardinal $$\kappa$$ and assume for a contradiction that $$\mathcal C$$ has the stated properties. For $$\alpha\in\kappa$$ let $$\mathcal C(\alpha)=\{c\in\mathcal C: \alpha\in c\}$$.

Claim 1. $$|\mathcal C|\ge\kappa$$.

Proof. The map $$\{c,d\}\mapsto c\cap d$$ is a surjection from $$\binom{\mathcal C}2$$ to $$\binom\kappa1$$.

Claim 2. $$\alpha\in\kappa\implies\mathcal C(\alpha)\ne\mathcal C$$,

Proof. Choose $$c\in\mathcal C(\alpha)$$, $$c\ne\{\alpha\}$$, choose $$\beta\in c\setminus\{\alpha\}$$, and choose $$d\in\mathcal C(\beta)\setminus\{c\}$$; then $$d\in\mathcal C\setminus\mathcal C(\alpha)$$.

Claim 3. $$\alpha\in\kappa,\ d\in\mathcal C\setminus\mathcal C(\alpha)\implies|\mathcal C(\alpha)|\le|d|\lt\kappa$$.

Proof. The map $$c\mapsto c\cap d$$ is an injection from $$\mathcal C(\alpha)$$ to $$\binom d1$$.

Now choose $$c\in\mathcal C$$, $$\alpha\in c$$, and $$d\in\mathcal C(\alpha)\setminus\{c\}$$. Let $$\lambda=\max(|d|,|\mathcal C(\alpha)|)\lt\kappa$$. Since $$|\mathcal C(\beta)|\le|d|$$ for all $$\beta\in c\setminus\{\alpha\}$$, and since $$\mathcal C=\bigcup_{\beta\in c}\mathcal C(\beta)$$, we have $$|\mathcal C|\le|c|\cdot\lambda\lt\kappa$$, contradicting Claim 1.