sequentially compact $d(x,y)=|\arctan(x)-\arctan(y)|$

Question

Let $\mathbb{R}$ equipped with the metric $\forall x,y \in \mathbb{R}$ $d(x,y)=|\arctan(x)-\arctan(y)|$.

Show that $\mathbb{R}$ is not sequentially compact with respect to that metric.

I know that it s bounded by $d(x,y)<\pi$.

But it should not be closed. But having trouble coming up with sequences that will converge but not in the metric space.

Also know that $x, y$ are unbounded so somehow the image will not be?

Would it be easier to use open covers, so for an arbitrary open cover there is no finite cover?

Answer 1

The sequence $1,2,3,4,\ldots$ is a Cauchy sequence with respect to that metric. But there is nothing in that space to which any of its subsequences could converge.

(To see that this is a Cauchy sequence, remember that $\arctan x\to\pi/2$ as $x\to\infty.$)