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arxiv: 1802.08843 · v3 · pith:RVUANX4Dnew · submitted 2018-02-24 · 🧮 math.CO

On the sizes of k-edge-maximal r-uniform hypergraphs

classification 🧮 math.CO
keywords hypergraphuniformedge-maximalsubhypergraphthenbestboundcalled
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Let $H=(V,E)$ be a hypergraph, where $V$ is a set of vertices and $E$ is a set of non-empty subsets of $V$ called edges. If all edges of $H$ have the same cardinality $r$, then $H$ is a $r$-uniform hypergraph; if $E$ consists of all $r$-subsets of $V$, then $H$ is a complete $r$-uniform hypergraph, denoted by $K_n^r$, where $n=|V|$. A hypergraph $H'=(V',E')$ is called a subhypergraph of $H=(V,E)$ if $V'\subseteq V$ and $E'\subseteq E$. A $r$-uniform hypergraph $H=(V,E)$ is $k$-edge-maximal if every subhypergraph of $H$ has edge-connectivity at most $k$, but for any edge $e\in E(K_n^r)\setminus E(H)$, $H+e$ contains at least one subhypergraph with edge-connectivity at least $k+1$. Let $k$ and $r$ be integers with $k\geq2$ and $r\geq2$, and let $t=t(k,r)$ be the largest integer such that $(^{t-1}_{r-1})\leq k$. That is, $t$ is the integer satisfies $(^{t-1}_{r-1})\leq k<(^{t}_{r-1})$. We prove that if $H$ is a $r$-uniform $k$-edge-maximal hypergraph such that $n=|V(H)|\geq t$, then ($i$) $|E(H)|\leq (^{t}_{r})+(n-t)k$, and this bound is best possible; ($ii$) $|E(H)|\geq (n-1)k -((t-1)k-(^{t}_{r}))\lfloor\frac{n}{t}\rfloor$, and this bound is best possible. This extends former results in [8] and [6].

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