I have the following problem:
Prove: the metrics $d$ and the metrics $2d$ in space $X$, where $(2d)(x,y) = 2d(x,y)$ are equivalent(i.e. induce the same topology).
Currently, I did the following sfuff:
- Let: $r\in \mathbb{R}^{+}$, $x\in X $ and let $B_1, B_2$ denotes balls in metrics space $(X,d), (X,2d)$ respectively.Assuming relationships between metrics $d$ and $2d$ and assuming the existence of the ball in metric space $(X,d)$, we have:
$B_1(x,r) = \{y\in X: d(x,y) < r\} = \{y\in X: \frac{1}{2}(2d)(x,y) < r \} = \{y\in X: (2d)(x,y) < 2r\} = B_2(x,2r)$
hence, the existence of ball $B_1$ in metric space $(X,d)$ implies equality with the ball $B_2$ in metric space $(X,2d)$
Analogically, assuming relationships between metrics $d$ and $2d$ and assuming the existence of the ball in metric $(X,2d)$, we have:
- $B_2(x,r) = \{y\in X: (2d)(x,y) < r\} = \{y\in X: d(x,y) < \frac{r}{2} \} = B_1(x,\frac{r}{2})$
- hence, the existence of ball $B_1$ in metric space $(X,d)$ implies equality with the ball $B_2$ in metric space $(X,2d)$
The above computation implies that $d$-ball is also $2d$-ball and that $2d-ball$ is also $d$-ball. This implies that the base (generating sets - open sets) are the same. Thus, the considered metrics are equivalent(they induce the same topology).
What else should I also contain in my proof to complete it?
Advices/suggestions very appreciated!