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Can someone help me with the following problem:

Prove that every open set in $\mathbb E^2$ is also open in $(\mathbb R^2, d_1)$, where $d_1((x_1,y_1),(x_2, y_2))=|x_1-x_2|+|y_1-y_2|$ , and vise versa, every open set in $(\mathbb R^2, d_1)$ is open in $\mathbb E^2$ also.

Kat
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    What have you tried so far? In finite-dimensional vector spaces all norms are equivalent, so in particular the 1 and 2 norms in $\mathbb R^2$ are equivalent. – Dave Jun 04 '19 at 21:22
  • Is the metric on $\mathbb{E}^2$ the Euclidean one? (square root of sum of squared differences) – coffeemath Jun 04 '19 at 21:23
  • Yes, it's the Euclidean metric – Kat Jun 04 '19 at 21:27

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Let $A$ be open in $\mathbb E^{2}$ and $(x,y) \in A$. There exists $r >0$ such that the open ball of radius $r$ around $(x,y)$ in $\mathbb E^{2}$ is contained in $A$. Let $(u,v)$ belong to the open ball around $(x,y)$ of radius $r/\sqrt2 $ in the $d_1$ metric. Then $|u-x| <r/\sqrt 2$ and $|v-y| <r/\sqrt 2$. Hence $\sqrt {(u-x)^{2}+(v-y)^{2}} <\sqrt {\frac {r^{2}} 2+\frac {r^{2}} 2}=r$, so $(u,v) \in A$.

The argument for the converse is similar. [Here you can use $r/2$ instead of $r /\sqrt 2$].

Kavi Rama Murthy
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