Let $X$ be a set. Two metrics $d_1, d_2: X\times X\to\mathbb{R}$ are equivalent, if constants $\alpha,\beta > 0$ exist such that for all $x,y\in X$ holds:
$\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$
Show that equivalent metrics generate the same topology
Proof:
Let $d_1, d_2: X\times X\to\mathbb{R}$ be metrics and $\tau_1, \tau_2$ the induced topolgies.
We have to show, that $\tau_1=\tau_2$.
- $\tau_1\subseteq\tau_2$.
Let $U\in\tau_1$ open. Then exists for every $x\in U$ a $\epsilon >0$ such that $B_{d_1}(x,\epsilon)\subseteq U$.
It is $B_{d_1}(x,\epsilon)=\{y\in X|d_1(x,y)<\epsilon\}$.
Since $d_1$ and $d_2$ are equivalent, there are constants $\alpha,\beta >0$ such that $\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$. Take $\epsilon':=\beta^{-1}\epsilon >0$.
We get $d_2(x,y)\leq \beta d_1(x,y)<\beta\cdot \beta^{-1}\cdot\epsilon=\epsilon$.
Hence $U\in\tau_2\checkmark$.
The other inclusion $\tau_1\supseteq\tau_2$ works analogously.
Is this proof correct? Thanks in advance.