Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an integer $\lambda$ such that $a_{\lambda}\subseteq m^{k}$. That is, the linear topology defined by $a_{\lambda}$ is finer than the topology defined by $m^{k}$.

Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...

Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an index $\lambda$ such that $a_{\lambda}\subseteq m^{k}$. That is, the linear topology defined by $a_{\lambda}$ is finer than the topology defined by $m^{k}$.

Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...

Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an integer $l$ such that $\bigcap_{\lambda} (m^{l}+a_{\lambda})\subseteq m^{k}$.

Could anyone give a proof of this statement or a counter example? Feel free to add some more mild assumptions...

Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an integer $\lambda$ such that $a_{\lambda}\subseteq m^{k}$. That is, the linear topology defined by $a_{\lambda}$ is finer than the topology defined by $m^{k}$.

Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...