Complete metric on set of rational numbers

Question

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!

Answer 1

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.