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Define $( \mathbb R^n, m) $ as a metric space such that $ m=\max \{ | x_i - y_i |: i=1...n \} .$

And also $(\mathbb R^2, d) $ as another metric space such that $d= \sum_{i=0}^n | x_i - y_i | .$

Question asks: are these metrics equivalent?

I know l should show the equivalent of induced topologies or $\tau_m $ = $ \tau_d $.

So l should pick an open ball from $ \tau_m$ and show that this open ball is open in $\tau_d $. But doesn't look like these basis balls are equivalent, as one of them is in $\mathbb R^2$ and the other is in $\mathbb R^n$ . Could you please evaluate my approach?

The second question is , what if l change the second metric to ( $\mathbb R^n$, d)?

Then $\mathcal B_\epsilon ^m (x) $ should be open in $\tau_d $.

In this case, what can l do to show these basis balls are open in both topologies?

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  • $\begingroup$ I cannot understand your second question: To which metric is $d$ changed? $\endgroup$ – Stephan Kulla Nov 9 '15 at 0:25
  • $\begingroup$ @tampis d is charged to ($\mathbb R^n $, d) $\endgroup$ – user288280 Nov 9 '15 at 0:43
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    $\begingroup$ One way to show equivalence of metrics $d1,d2$: For a point $p$ and $\{i,j\}=\{1,2\}$ and $r>0$ show there exists $s>0$ such that $B_{dj}(p,s)\subset B_{di}(p,r)$. $\endgroup$ – DanielWainfleet Nov 9 '15 at 0:45
  • $\begingroup$ @user288280 What is the definition of $d$ in the second question? From the word "change" I deduce that it is not the metric $d$ defined in your first sentence... $\endgroup$ – Stephan Kulla Nov 9 '15 at 0:49
  • $\begingroup$ @tampis l just edited the second question ... sorry for that $\endgroup$ – user288280 Nov 9 '15 at 0:52
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For the first question: $(\mathbb R^n, m)$ and $(\mathbb R^2, d)$ are only be equivalent metric spaces iff $n=2$ because the underlying vector space need to be the same. For $n=2$ see the following paragraph...

For the second question: Your metrics are defined by p-norms and because your vector space is real-valued and finite the metrics are equivalent. See Any two norms equivalent on a finite dimensional norm linear space. why this is the case.

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  • $\begingroup$ Thanks. .. is there any other way to show it by open balls on both metric spaces for second question? I mean not using the p - norms $\endgroup$ – user288280 Nov 9 '15 at 1:02
  • $\begingroup$ Yes, but I would recommend to show that the underlying norms are equivalent... $\endgroup$ – Stephan Kulla Nov 9 '15 at 1:06

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