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Prove that any subset of $\mathbb{R}^2$ which is $D$-open is $D_1$-open and, conversely, that any subset of the plane which is $D_1$-open is $D$-open.

Let $D$ be $D(x_1,y_1)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Let $D_1$ be $D((x_1,y_1)(x_2,y_2))=\lvert x_1 - x_2 \rvert + \lvert y_1-y_2 \rvert$.

Attempt:

Consider the metric space $(X,D)$. Let $U\subset X$. Then, for any point $(x,y)\in U$ there exists $p>0$ where $N_p(x,y)\subset U$.

Consider the metric space $(X,D_1)$. Let $V\subset X$. Then, for any point $(x_1,y_1)\in V$ there exists $q>0$ where $N_q(x_1,y_1)\subset V$.

Given $p$, let $q=p/\sqrt{2}$. Then for any $(x',y')\in N_q(x)\subset V$, $(x',y')\in N_p(x)\subset U$. Hence, if $X$ is $D$-open, then it must also be $D_1$-open.

Any help fixing this proof is appreciated.

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    $\begingroup$ It would help if you told us what $D$ and $D_1$ are. $\endgroup$ – Mr. Chip Sep 26 '13 at 11:15
  • $\begingroup$ Okay I have updated the question. Thank you. $\endgroup$ – Julius Jackson Sep 26 '13 at 19:31
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$D$ is the usual euclidean metric; so it is translation invariant. The $D$-disk of radius $r>0$ with center ${\bf 0}=(0,0)$ is the usual circular disk with radius $r$.

The metric $D_1$ is translation invariant as well. The $D_1$-disk of radius $\rho>0$ with center ${\bf 0}$ consists of the points $(x,y)\in{\mathbb R}^2$ with $|x|+|y|<\rho$, and is a square $Q_\rho({\bf 0})$ with $45^\circ$ sides centered at ${\bf 0}$.

Given a $D$-open set $\Omega\subset{\mathbb R}^2$ each point ${\bf p}\in\Omega$ is the center of a $D$-disk $B_r({\bf p})\subset\Omega$, and this disk $B_r({\bf p})$ contains $Q_r({\bf p})$. It follows that $\Omega$ is $D_1$-open as well.

Similarly: Given a $D_1$-open set $\Omega\subset{\mathbb R}^2$ each point ${\bf p}\in\Omega$ is the center of a $D_1$-disk $Q_\rho({\bf p})\subset\Omega$, and this disk $Q_\rho({\bf p})$ contains $B_r({\bf p})$, where $r=\rho/\sqrt{2}$. It follows that $\Omega$ is $D$-open as well.

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You have to specify the two metrics. Otherwise you can't get any help.
For example if we have the two following metrics over $\mathbb{R}^2$:
$d:\mathbb{R}^2 \rightarrow \mathbb{R}$ the standard euclidean metric and
$d_1:\mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $d_1(x,x)=0$ and $d_1(x,y)=1$ if $x\neq y$
then your statement isn't true.

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  • $\begingroup$ I think D1 is referred to as the "taxi-cab" metric. $\endgroup$ – Julius Jackson Sep 26 '13 at 18:07

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