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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?

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  • $\begingroup$ 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 $d_2$. $\endgroup$
    – Mirko
    Nov 19, 2019 at 14:17
  • $\begingroup$ 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. $\endgroup$
    – Moe1234
    Nov 19, 2019 at 14:23

1 Answer 1

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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 $(\mathbb R,d_1)$.

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  • $\begingroup$ I know what a homeomorphism is, but it is never defined in my script. If i have to show, that two metrics are equivalent, i propably am not allowed to use this way. Is there an other way to show this? $\endgroup$
    – Moe1234
    Nov 19, 2019 at 14:04
  • $\begingroup$ 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? $\endgroup$ Nov 19, 2019 at 14:35
  • $\begingroup$ Yes. I think i can use that. Thanks $\endgroup$
    – Moe1234
    Nov 19, 2019 at 15:08

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