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A linear hypergraph is a hypergraph $H=(V,E)$ such that

  1. $|e|\geq 2$ for all $e\in E$,
  2. $|e_1\cap e_2|\leq 1$ for all $e_1, e_2\in E$ with $e_1\neq e_2$.

We call a linear hypergraph complete if there is equality in statement 2 above, i.e. if $|e_1\cap e_2| = 1$ for all $e_1, e_2\in E$ with $e_1\neq e_2$.

Moreover, a linear hypergraph is saturated if for all $v\in V$ we have $|\{e\in E:v\in e\}| \geq 2.$

For $n>2$ set $\mathbb{N}_n =\{1,\ldots,n\}$ and let $$\ell_1(n)=\min\{|E|: E\subseteq{\cal P}(\mathbb{N}_n) \text{ and }(\mathbb{N}_n, E) \text{ is linear, saturated}\},$$ and $$\ell_2(n)=\min\{|E|: E\subseteq{\cal P}(\mathbb{N}_n) \text{ and }(\mathbb{N}_n, E) \text{ is linear, saturated & complete}\}.$$

Obviously, we have $\ell_1(n) \leq \ell_2(n)$, but do we have equality?

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1 Answer 1

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You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where any $r$ edges share at most $1$ point and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

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