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?
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 $(\mathbb R,d_1)$.