Let $(R,\mathfrak{m})$ be a Noetherian local ring and let $(I_{n})_{n\in\mathbb{N}}$ be a decreasing chain of ideals in $R$ such that $\bigcap_{n \in \mathbb{N}} I_{n} = \{ 0 \}$. Then there is a unique separated topology $\tau$ on $R$ which has the family $(I_{n})_{n\in \mathbb{N}}$ as a zero neighborhood and which turns $R$ into a topological ring. If $I_{n} = \mathfrak{m}^{n}$ then it is known that the completion of $R$ has the same dimension as $R$ and in fact this holds also if $I_{n} = J^{n}$ for any proper ideal $J$ of $R$ (see Krull dimension of a completion ).
Does it hold in general for any chain of ideals $(I_{n})_{n \in \mathbb{N}}$ that the completion of $R$ wrt to the uniformity defined by $(I_{n})_{n \in \mathbb{N}}$ has the same dimension as $R$?
Note that if $R$ is $\mathfrak{m}$-adically complete then it is complete in any separated linear topology (any topology obtained in the above way).