Are d1 and lift metric equivalent distances?

Question

I need help proving that $d1\not\equiv d$, where d and d1 are defined as follows: \begin{equation} \label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande} d(x,y) = \left\{ \begin{array}{ll} |x_{2}-y_{2}| & \mathrm{if \ }x_{1}=y_{1} \\ |x_{2}|+|x_{1}-y_{1}|+|y_{2}| & \mathrm{otherwise\ } \end{array} \right. \end{equation}

and $d_1=|x_1-y_1|+|x_2-y_2|$

$x,y\in \mathbb{R}^2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$,

I have that

$d_1(x,y)\leq d(x,y) $, $B_d(x;\epsilon)\subseteq B_{d_1}(x;\epsilon)$ and $\tau_u\subseteq \tau_d$ where $\tau_u$ is the usual topology induced by $d_1,d_2,d_\infty$ on $\mathbb{R}^2$and $\tau_d$ the topology induced by d on $\mathbb{R}^2$.

I need to prove that HOWEVER $d_1$ and $d$ are not equivalent.

Thanks.

Answer 1

$D=\{0\}\times \{y:1<y<3\}$ is the open $d$-ball of radius $1$ centered at the point $(0,2).$ It is not open in the $d_1$-metric.

For any $p=(x,y)\in \Bbb R^2$ and any $r>0$ the open $d_1$-ball $B_{d_1}(p,r)$ contains points $(x',y')$ with $x'\ne x. \;$ E.g. $(x+r/2,y)\in B_{d_1}(p,r). $ So no non-empty open $d_1$-ball is a subset of $D.$ And $D$ is not empty. So $D$ cannot be a union of open $d_1$-balls.

The metric $d$ has been called the River Metric (e.g. in General Topology by R. Engelking): For real $x,x'$ with $x\ne x',$ the sets $\{x\}\times \Bbb R$ and $\{x'\}\times \Bbb R$ are separated by mountains, except for the river, which is $\Bbb R\times \{0\}.$ To get from $(x,y)$ to $(x',y')$ you must travel to $(x,0)$ and along the river to $(x',0)$ and then to $(x',y').$

Answer 2

$d_1 ((\frac 1 n , 1+ \frac 1 n ) ,(0,1))\to 0$ but $d_2((\frac 1 n , 1+ \frac 1 n ),(0,1)$ does not approach 0. There are several definitions of equivalent metrics and one of them says convergent sequences and their limits coincide for the two metrics. This is the definition I am using. If you have a different definition I can show you how to prove the equivalence.

Answer 3

In the lift metric (or river metric, as I know it) points on the $x$-axis do have equivalent neighbourhood systems, but off the $x$-axis, the neighbourhoods are vertically oriented. For $y \neq 0$, $B_d(x,y), |y| ) = \{x\} \times (0,2y)$, e.g.

This last type of ball clearly does not contain any $d_1$-ball in the plane. So the metrics are not equivalent, which can also be shown by differing convergence properties: under $d_1$ $(\frac{1}{n}, 1) \to (0,1)$ but under $d$ we have $d((\frac{1}{n}, 1), (0,1)) = 2 + \frac{1}{n}$ so that $(\frac{1}{n},1) \not\to (0,1)$ under $d$.