Metric with $\arctan$ defines the same topology as the Euclidean metric on $\mathbb{R}$

Good day,

I have the following exercise: Show that the metric $d_{\arctan}(x,y):=|\arctan(x)-\arctan(y)|$ defines the same topology as $d_{id}=|x-y|$.

I have the following theorem to use: Let $d_1, d_2$ be two metrices on $X$. They are topologically equivalent if there are $k_1, k_2 >0$ such that $\forall x,y \in X : k_1 d_2(x,y) \leq d_1(x,y) \leq k_2 d_2(x,y)$

So I have to show that there are $k_1,k_2 > 0$ s.t. $\forall x,y \in \mathbb{R} : k_1 |x-y| \leq |\arctan(x)-\arctan(y)| \leq k_2 |x-y|$

The right inequality should follow easily by the mean value theorem. But for the left side I have no clue. I think I have to use that $\arctan(x) \in (-\frac{\pi}{2}, \frac{\pi}{2} ) ~\forall x\in \mathbb{R}$

Maybe some help?

Thanks a lot,

Marvin

• There is no $k_1$ so that the left inequality holds for all $x, y$. I therefore think you should first switch from the Euclidean metric to the following, topologically equivalent, metric: $$d(x, y) = \min(|x-y|, 1)$$ Alternatively, you can use another theorem: Two topologies are equivalent if every basic open set (i.e. open balls) in one topology is an open set in the other. Oct 27 '15 at 9:08
• The if goes one way: if such constants exists then they are topologically equivalent. But then need not exist (and this is an example of this!). Oct 27 '15 at 9:26
• Yeah, I realized this, but this was the only theorem for topologically equivalence we had in the lecture so I hoped I can use it here. But Arthur is right, I can also use the property of basic open sets, it was also used in the proof to the theorem. I have to think about this, how this works in my case. Thanks you two. Oct 27 '15 at 9:46

Indeed, you do not have such an inequality. The metric induced by $$|\arctan x - \arctan y|$$ is bounded. So the left inequality cannot be attained.
$$\arctan : \mathbb R \to (-\pi/2, \pi/2)$$
is a homeomorphism and the metric $|\arctan x - \arctan y|$ is just the pullback of the standard one on $(-\pi/2, \pi/2)$.