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?
Thanks in advance.
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.