# Partitioning finite hypergraphs with edges [closed]

Let $H=(V,E)$ be a hypergraph such that $|V|$ is infinite, and the following statements hold:

1. if $a\neq b\in E$ then $|a\cap b|\leq 1$, and
2. every vertex $v\in V$ is contained in at least $2$ members of $E$.

Is there $P\subseteq E$ such that the members of $P$ are pairwise disjoint, and $\bigcup P = V$? (This would amount to a kind of matching in hypergraphs.)

## closed as off-topic by Ben Barber, Jan-Christoph Schlage-Puchta, RP_, Chris Godsil, David HandelmanDec 6 '17 at 3:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ben Barber, Jan-Christoph Schlage-Puchta, RP_, David Handelman
If this question can be reworded to fit the rules in the help center, please edit the question.

• What if $H$ is a graph which is the union of infinitely many vertex-disjoint triangles? – bof Dec 1 '17 at 6:58
• In that case, each $v\in V$ is contained only in one member of $E$, or do I misunderstand something? – Dominic van der Zypen Dec 1 '17 at 7:14
• No doubt I misunderstand something. My $H$ is the union of infinitely many vertex-disjoint copies of the graph $K_3.$ Each vertex is in two edges, each edge contains two vertices, and two distinct edges have at most one vertex in common. – bof Dec 1 '17 at 9:22
• The edges don't have to be finite sets, do they? In the first example I thought of, the vertices are the points of the real projective plane, the points are the lines; two edges intersect in exactly one vertex, each vertex belongs to continuum many edges. – bof Dec 1 '17 at 9:27

Consider the following (connected) hypergraph: i.e. $V = \{v,w,x\}\cup\{1,2,3,\dots\}, E = \{\color{blue}{\{v,x\}}, \color{red}{\{v,w\}}, \color{orange}{\{w,x,1\}}\}\cup\{\{1,2\}, \{2,3\}, \dots\}$
To cover the vertex $v$ you need to include the blue or the red edge in $P$. But then you can't also include the yellow edge. Thus either $x$ or $w$ won't be covered.