# Equivalent metrics and Cauchy sequences

I know, that two metrics $$d_1$$ and $$d_2$$ on $$X$$ are equivalent, if every set, that is open regarding $$d_1$$, is also open regarding $$d_2$$. But i have no idea how to show that.

Let $$d_1$$ be the standard metric on $$\mathbb{R}$$ ($$d_1(x, y)=|x-y|$$), and let $$d_2=d_a$$, with $$d_a(x,y)=|arctan(x)-arctan(y)|$$ be a metric on $$\mathbb{R}$$.

Can someone show me how to show, that these two metrics are equivalent? (Without using homeomorphisms)

I know that for two equivalent metrics $$d_1$$ and $$d_2$$ on $$X$$, every sequence in $$X$$ converges regarding $$d_1$$, iff it converges regarding $$d_2$$.
But that propably doesn’t mean, that every Cauchy sequence regarding $$d_1$$ is also a Cauchy sequence regarding $$d_2$$, right?

Last question: Is „regarding“ used correctly?

• So, you are saying that you have no idea how to show that you know, that two metrics $d_1$ and $d_2$ on $X$ are equivalent, if every set, that is open regarding $d_1$, is also open regarding $d_2$. Nov 19, 2019 at 14:17
• Yes. I‘m not able to show, that every set, that is open regarding $d_1$ is open regarding $d_2$. I guess it‘s easy, but i don‘t know how. Nov 19, 2019 at 14:23

The map $$\arctan\colon\mathbb R\longrightarrow\left(-\frac\pi2,\frac\pi2\right)$$ is a homeomorphism. This is equivalent to the assertion that $$\operatorname{Id}\colon(\mathbb R,d_1)\longrightarrow(\mathbb R,d_2)$$ is a homeomorphism, which in turn is equivalent to the assertion that the metrics $$d_1$$ and $$d_2$$ are qeuivalent.
On the other hand, note that the sequence $$1,2,3,\ldots$$ is a Cauchy sequence in $$(\mathbb R,d_2)$$, but not in $$(\mathbb R,d_1)$$.
• Asserting that the metrics are equivalent is the same thing as asserting that the identity map is a homeomorphism. Now, think about how $d_2$ is defined. Doesn't that mean thar $\arctan$ is a homeomorphism? Nov 19, 2019 at 14:35