Let $A,B\neq \emptyset$ be disjoint and suppose $G = (A\cup B, E)$ is bipartite where for all $e\in E$ we have $e\cap A \neq \emptyset\neq e\cap B$. For $a\in A$ we set $N\_G(a) = \{b\in B: (\exists e\in E)\{a,b\}\in e\}$, and for $S\subseteq A$ let $N\_G(S) = \bigcup\{N\_G(a):a\in S\}$. 


A *matching* is a set $M$ of pairwise disjoint edges, and a *marriage* of $A$ is a matching such that $A\subseteq \bigcup M$. For finite bipartite graphs, [Hall's marriage theorem](https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem) says that there is a marriage if



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> (Hall's condition:) for all $S\subseteq A$ we have $|S| \leq |N\_G(S)|$.
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**Question**. Assuming that we $|A|=|B| = \aleph\_0$, is there $n\in\mathbb{N}$ such that: 



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> if $|N\_G(a)| \geq n$ for all $a\in A$ and Hall's condition holds for $G$, then $G$ has a marriage
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?




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**Background**: For $n=1$, the implication in the question fails: Let $A = \omega \times\{0\}$ and $B = \omega \times\{1\}$. Set $$E = \big\{\{(k,0), (k-1,1)\}: k\in\omega\setminus\{0\}\big\}\cup\big\{\{(0,0),(n,1)\}: n\in\omega\big\}.$$ Then it is easy to see that $G= (A\cup B,E)$ obeys Hall's condition, but $G$ has no marriage.



Marriages in infinite bipartite graphs with many neighbors
No. You may modify your example. Join $(k,0)$ with $(m,1)$ for all $m<k$ and also join $(k,0)$ with all $(m,1)$ for $k\leqslant n$. Then any matching must match $\{0,\dots,N\}\times 0$ with $\{0,\dots,N-1\}\times 1$ for large enough $N$. This is impossible.


