# Proof that $\mathbb{R}^2$ is complete with the metric $d(x,y) = \min\{\|x-y\|, 1\}$.

The question is as in the title: Proof that $\mathbb{R}^2$ is complete with the metric $$d(x,y) = \min\{\|x-y\|, 1\}.$$

One way of doing this is proving that the metric is topological equivalent with the normal metric in $\mathbb{R}^2$ and because $\mathbb{R}^2$ is complete with the normal metric, it follows that $\mathbb{R}^2$ is complete with this new metric. But I don't know how to show this metric is topological equivalent with the normal metric.

Or is there maybe a direct proof of the completeness of this metric?

• "One way of doing this is proving that the metric is topological equivalent with the normal metric": no, it isn't. Completeness of a metric space is not a topological invariant. For instance, Euclidean topology on $\Bbb R$ can be metrized both by $d(x,y)=\lvert x-y\rvert$ and by $d'(x,y)=\lvert \arctan x-\arctan y\rvert$. Yet, $(\Bbb R,d)$ is complete, while $(\Bbb R,d')$ isn't. – Saucy O'Path Jun 21 '18 at 9:43
• Ok, good to know, and $d(x,y)$ is topologically equivalent with $d'(x,y)$? – Belgium_Physics Jun 21 '18 at 10:01
• @Belgium_Physics, yes, it is; $\arctan : \mathbb R \to (-\pi, \pi)$ is a homeomorphism (with inverse $\tan$, of course). – Mees de Vries Jun 21 '18 at 10:03
• @Belgium_Physics Yes. Since $B_{d'}(x,\varepsilon)=\arctan^{-1}\left[B_{d}(\arctan x,\varepsilon)\right]$, a set is open in $(\Bbb R,d')$ if and only if it is the preimage by $\arctan$ of some open set $\Omega\subseteq (-\pi/2,\pi/2)$. As $\arctan:(\Bbb R,d)\to ((-\pi/2,\pi/2),d)$ is a homeomorphism, such sets are exactly the open sets of $(\Bbb R,d)$. – Saucy O'Path Jun 21 '18 at 11:35
If $(x_n)_{n\in\mathbb N}$ is a sequence of elements of $\mathbb{R}^2$, then it is a Cauchy sequence with respect to this metric if and only if it is a Cauchy sequence with respect to the usual metric. And, if $x\in\mathbb{R}^2$, $\lim_{n\to\infty}x_n=x$ with respect to this metric if and only if $\lim_{n\to\infty}x_n=x$ with respect to the usual metric. This is so because, if $\varepsilon<1$, asserting that $d(x,y)<\varepsilon$ means the same thing for both metrics. Since $\mathbb{R}^2$ is complete with respect to the usual metric, it is also complete with respect to this one.