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I know of no necessary conditions for a metric d to be equivalent to the standard euclidean metric on Reals.Hence, I was facing difficulty in answering the following problem: Does there exist d, equivalent to Standard metric on reals, s.t. it makes (Q,d) a complete metric space?

user168826
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    What do you know about equivalent metrics? What definition are you using? Do you know that two metrics are equivalent implies that any convergent sequence in one metric is a convergent sequence in the other? Because that would answer your question. – Thomas Andrews Feb 16 '15 at 19:32
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    The answers to this question contain proofs that a complete metric space with no isolated points is necessarily uncountable (and in fact necessarily of cardinality at least $2^\omega=\mathfrak{c}$). Thus, there is no such $d$, even if you don’t insist that it be the restriction of a metric on $\Bbb R$. – Brian M. Scott Feb 16 '15 at 19:39
  • Normally when one says two metrics are equivalent, it means each distance function is bounded by a (positive) constant multiple of the other distance function. There is a weaker notion of equivalence between two metrics, that they give the same topology. Even with this weaker notion, convergence of a sequence is preserved, so think through what Thomas Andrews has already Commented. – hardmath Feb 16 '15 at 19:42
  • OK, I get it!Thanks a lot. – user168826 Feb 16 '15 at 19:46
  • @hardmath: I don't see how the mere fact "convergence of a sequence is preserved" is very helpful in proving that $\mathbb Q$ does not have a complete metric which is "weakly equivalent" to the usual metric. After all, the space of real numbers, likewise the space of irrational numbers, have complete and incomplete metrics which induce the same topology. – bof Feb 16 '15 at 19:58
  • @ThomasAndrews: How does that answer the question? The convergent sequences are the same, but the Cauchy sequences are not the same. – bof Feb 16 '15 at 19:59
  • Every convergent sequence must be Cauchy. So there are some Cauchy sequences which are not convergent, but not visa versa. @bof In particular, once you have a convergent sequence of rationals in the usual metric, it must be convergent (to the same value) in the new metric, so it must be Cauchy in the new metric, but it has no limit in $\mathbb Q$.This is essentially Christian's answer. – Thomas Andrews Feb 16 '15 at 20:00
  • @ThomasAndrews: Oops. Somehow I overlooked that the OP wants to metric to be compatible with the topology Of $\mathbb R$, not just $\mathbb Q$. Of course that makes the question trivial, since a complete subspace of a metric space is closed. – bof Feb 17 '15 at 05:57

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No.

Proof. Let $(x_n)_{n\geq0}$ be a sequence of rational numbers converging to $\sqrt{2}$ with repect to the standard metric in ${\mathbb R}$, and let $d:=d(\cdot,\cdot)$ be some other metric on ${\mathbb R}$, equivalent to the standard metric. Then $\lim_{n\to\infty} x_n=\sqrt{2}$ also with respect to $d$, whence $(x_n)_{n\geq0}$ is a Cauchy sequence with respect to $d$. If $({\mathbb Q},d)$ were complete then $(x_n)_{n\geq0}$, being a sequence of rational numbers, would have to converge to some $\xi\in{\mathbb Q}$ with respect to $d$, but this would imply $\sqrt{2}\in{\mathbb Q}$ – a contradiction.

Christian Blatter
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Here is some different aproach,

$\mathbb{Q}$ is countable so we can enumerate $\mathbb{Q}$ as $\{x_1,x_2,x_3,...\}$.

Suppose there exist a metric $d$ equivalent to the standard metric such that it makes $(\mathbb{Q},d)$ a complete metric space.

Now, $\mathbb{Q}=\cup_{n=1}^{\infty}\{x_n\}$

$\{x_n\}$ is a finite set so it is closed in $\mathbb{Q}$ with respect to the standard metric as well as the $d$ metric.

Let, $\{x_n\}=A_n$ , $\forall n \in \mathbb{N}$

here , ${(A_n)}^{o}=\phi$ , i.e, the interior of the set $A_n$ is $\phi$ for all $n \in \mathbb{N}$.Since ${(A_n)}^{o}=\phi$ in $\mathbb{Q}$ with respect to the standard metric, which implies that it will be same in $(\mathbb{Q},d)$ since $d$ is equivalent to the standard metric.

Now according to $\textbf{Baire Category Theorem}$ ,

If $(X,d)$ be a complete metric space and $\{F_n\}_n$ be a sequence of non empty closed subsets of $X$ such that $X=\cup_{n=1}^{\infty}F_n$ . Then at least one of $F_n$'s has nonempty interior.In other words,a complete metric space cannot be a countable union of nowhere dense closed subsets.

nowhere dense set: A subset of a metric space $X$ is called nowhere dense in $X$ if the interior of the closure of the subset is empty. i.e., $(\bar{A})^{o}=\phi$.

Now $\{A_n\}_n$ is a countable sequence of nonempty closed subsets of $\mathbb{Q}$ and all $A_n$'s are nowhere dense subsets of $\mathbb{Q}$ and also $\mathbb{Q}$ can be written as countable union of nowhere dense subsets.

So it creates a contradiction since we assume that $(\mathbb{Q},d)$ is a complete metric space.

So there does not exist such metric $d$ which makes $(\mathbb{Q},d)$ a complete metric space where $d$ is equivalent to the standard metric.

here is a link https://en.wikipedia.org/wiki/Baire_category_theorem.

Supriyo Banerjee
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