Can someone solve it? Same for $\mathbb{Q} \times\mathbb{Q}$ in $d(a,b) = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2}$ metric


1 Answer 1


I don't think this is true. For example consider the Cauchy sequence

$$\left\{(3,0), (3.1, 0), (3.14, 0), (3.141, 0), \dots \right\} $$

where each step adds another digit of $\pi$. This is a Cauchy sequence that will converge to $(\pi, 0)$, which is not in $\mathbb{Q} \times \mathbb{R}$.

  • 2
    $\begingroup$ (+1) More generally, this $d$ is just the usual Euclidean metric on the subspace $\Bbb Q\times\Bbb R$ of $\Bbb R^2$. A subspace $Y$ of a complete metric space $X$ is complete with respect to the same metric if and only if $Y$ is a closed set in $X$. $\Bbb R^2$ is complete with respect to the Euclidean metric, and $\Bbb Q\times\Bbb R$ is not closed in $\Bbb R^2$, so $\Bbb Q\times\Bbb R$ is not complete with respect to $d$. $\endgroup$ Commented Jan 19, 2021 at 21:14

Not the answer you're looking for? Browse other questions tagged .