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,


  • 1
    $\begingroup$ 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. $\endgroup$
    – Arthur
    Oct 27, 2015 at 9:08
  • 2
    $\begingroup$ 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!). $\endgroup$ Oct 27, 2015 at 9:26
  • $\begingroup$ 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. $\endgroup$ Oct 27, 2015 at 9:46

1 Answer 1


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.

Indeed, you do not need to use that. The mapping

$$\arctan : \mathbb R \to (-\pi/2, \pi/2)$$

is a homeomorphism and the metric $|\arctan x - \arctan y|$ is just the pullback of the standard one on $(-\pi/2, \pi/2)$.

I am a bit skeptical about that term "topologically equivalent". It looks like a metric equivalence instead.

  • 2
    $\begingroup$ Indeed, this is metric equivalence (a stronger property, which implies that the metric spaces have the same uniform properties, so the same Cauchy sequences etc.). Metric equivalence implies topologically equivalent (i.e. having the same open sets), but not vice versa, as this standard example shows, in fact. $\endgroup$ Oct 27, 2015 at 9:25

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