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Let $G=(S,T;E)$ be a bipartite graph without isolated vertices.

For every edge $e\in E$, e $=$ $st$ $($ s $\in S$, $ t\in T$) happens the inequality $dG(s)$ $>=$ $dG(t)$.

Prove that in $G$ exists a matching which saturates all the vertices of S.

My thoughts:

I found that there exists $Theorem$ $Hall(1935)$ which says:

Let $G$ $=$ $(S,T;E)$ be a bipartite graph. There exists a matching in $G$ which satures all the vertices in $S$ if only and only

$|NG(A)|$ $>=$ $|A$|, $∀A ⊆ S$. (NG represents the neighbors i suppose)

But i don't know how to adapt Hall's Theorem for my property that i receive $dG(s)$ $>=$ $dG(t)$.

Would be glad for some answers or other ideas to proof that there exists a matching with that relation.

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1 Answer 1

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This is classical. Take arbitrary non-empty subset $A\subset S$. Consider the sum of $1/dG(s)$ over all edges $st\in E$, $s\in A$. On the one hand, it equals $|A|$. On the other hand, $\sum_{s:st\in E,s\in A} 1/dG(s)\leqslant \sum_{s:st\in E,s\in A} 1/dG(t)\leqslant |NG(A)|$, thus $|A|\leqslant |NG(A)|$ as desired.

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