Does there exist a metric on $\mathbb{Q}$ which is equivalent
to the standard metric but $(\mathbb{Q}, d)$ is complete.??
My attempt :
We know that with respect to standard metric, each singleton is closed subsets.
Since we know that a countable union of nowhere dense sets in a metric space need not be nowhere dense.
As for example, consider the set of rational numbers as a subset of $\mathbb{R}$, we can write $\mathbb{Q}$ as the union of singleton set, which of course nowhere dense sets in $\mathbb{R}$.
But, of closure ($\mathbb{Q}$) is $\mathbb{R}$ and $\mathbb{Q}$ is everywhere dense in $\mathbb{R}$ and hence, $\mathbb{Q}$ is not nowhere dense set in $\mathbb{R}$. Therefore, there exist a metric s. t $(\mathbb{Q}, d)$ is complete.
Is my arguments true.??
Please help me. Thanks
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $\Bbb Q$ is countable.
Also your argument does not actually establish a metric on $\Bbb Q$ that makes $\Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.