Does there exist a metric on Q which is equivalent to the standard metric but ( Q, d) is complete?
We know that with respect to standard metric, each singleton is a closed subsets.
And A countable union of nowhere dense sets in a metric space need not be a nowhere dense set.
For example set of rational Q as a subset of R, is countable union of singleton 's which of course are nowhere dense set . But closure of rational is R and such Q is everywhere dense in R and Q is not nowhere dense set in R.
So, I think the above argument help to answer my question. Please help. Thanks!
No such metric exists because every complete metric space without isolated points is either uncountable or empty.
The proof is similar to what you wrote. Suppose $X$ is such a space. Since there are no isolated points, $X\setminus\{a\}$ is a dense open subset of $X$ for each $a\in X$. If $X$ is at most countable, then the fact that $$ \bigcap_{a\in X} X\setminus \{a\} = \varnothing $$ contradicts the Baire category theorem, according to which a countable intersection of dense open subsets of a nonempty complete metric space is nonempty.