All our rings are commutative with unity.


For an $R$-module $M$ and a submodule $N$ of $M$ and ideal $I$ of $R$, let $(N:I):=\{m\in M : Im \subseteq N\}$. Let $\mu (M)$ denote the least cardinality among the generating sets of $M$. 


Now let $\alpha$ be an infinite cardinal. Let $M$ be a faithful $R$-module such that $\mu(M) <\alpha$. If $r\in R ,m \in M$ and $N$ is a submodule of $M$ such that $rm \in N$ and $\mu (N+Rm)<\alpha$ and $\mu ((N: Rr)) <\alpha$, then is it true that $\mu (N) <\alpha$ ? If this is not true for every infinite cardinal $\alpha$, then for which $\alpha$ is it true ? 


I can show that the assertion is true when $M=R$. 



On minimal generating sets of certain submodules
Are you missing some conditions? I think the following is a counterexample with $M=R$.


Let $k$ be a field and $R=k[x\_i\mid i\in I]$, with $|I|=\alpha$, a polynomial ring in $\alpha$ many variables. 


Take $M=R$, $N=\langle x\_i\mid i\in I\rangle$, $m=1$, and $r$ any element of $N$.


Then 


* $\mu(M)=1$,
* $\mu(N)=\alpha$,
* $rm=r\in N$
* $N+Rm=R$, so $\mu(N+Rm)=1$,
* $(N:Rr)=R$, so $\mu\left((N:Rr)\right)=1$.


