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Let $A$ be a subset of $\mathbb{R}^2$. Prove that $A$ is open in $(\mathbb{R}^2,d_0) \iff A$ is open in $(\mathbb{R}^2,d_1)\iff A$ is open in $(\mathbb{R}^2,d_2)$, where $$d_0((x_1,y_1),(x_2,y_2))=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\;,$$ $$d_1((x_1,y_1),(x_2,y_2))=\lvert x_1-x_2\rvert + \lvert y_1-y_2\rvert\;,$$ and $$d_2((x_1,y_1),(x_2,y_2))=\max\{\lvert x_1-x_2\rvert, \lvert y_1-y_2\rvert\}\;.$$


I think I have to just show that $d_0\cong d_1\cong d_2$.

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  • $\begingroup$ Right. If the metrics are equivalent, they induce the same topology (by definition), and openness is a topological property. $\endgroup$ – Daniel Fischer Sep 3 '13 at 21:45
  • $\begingroup$ Those are norms and all norms are equivalent on a finite dimensional vector space ... $\endgroup$ – Dominic Michaelis Sep 3 '13 at 21:45
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    $\begingroup$ Note that $d_2\le d_0\le d_1$. This gives you immediately, that the topology generated by $d_1$ is finer than that for $d_0$, which is finer than that for $d_2$. So the main difficulty is in showing the reverse inclusions. $\endgroup$ – Stefan Hamcke Sep 3 '13 at 21:47
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Hint: By Stefan's comment it remains to find constants $c, k>0$ such that $d_1\leq cd_0\leq kd_2$. A hint for doing this is for each $i$ draw the $\epsilon$ ball with respect to the metric of $d_i$ for some $\epsilon$ (say $\epsilon=1$).

See how they fit together and try to work this into a proof.

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