Rational Number Space is not Complete Metric Space

Theorem
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean metric $d$.

Then $\struct {\Q, \tau_d}$ is not a complete metric space.

Proof
Proof by Counterexample:

First note that Rational Numbers form Metric Space.

It remains to be shown that $\struct {\Q, \tau_d}$ is not complete.

Consider the sequence $\sequence {a_n}$ given by:
 * $a_n := \dfrac {f_{n + 1} } {f_n}$

where $\sequence {f_n}$ is the sequence of Fibonacci numbers.

By Ratio of Consecutive Fibonacci Numbers:
 * $\ds \lim_{n \mathop \to \infty} a_n = \phi := \dfrac {1 + \sqrt 5} 2$

But Square Root of Prime is Irrational.

That means $\dfrac {1 + \sqrt 5} 2$ is likewise irrational.

Thus $\sequence {a_n}$ is a Cauchy sequence of rational numbers which converges to a number which is not in $\Q$.

The result follows by definition of complete metric space.

In fact, any sequence of rational numbers that converges to an irrational number (in the metric space $\R$) is a Cauchy sequence that does not converge in $\Q$.

Also see

 * Normed Vector Space of Rational Numbers is not Banach Space