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Two metrics $d_1$ and $d_2$ are called equivalent if there exist positive constants $\alpha, \beta$ s.t $\forall x,y\in\mathbb R^n: \alpha d_2(x,y)\le d_1(x,y)\le\beta d_2(x,y)$

I already proved that $d_1$=Manhattan Metric and $d_\infty$ are equivalent.

I do not see the equivalence between $d_\infty=\max |x_y-y_j|$ and $d_2=\sqrt{\sum_{j=1}^{n}(x_j-y_j)^2}$, may you could help me.

2nd Question: The infinite intersection of open sets do not have to be open: Of youre I know the well known example $\bigcap(-1/n,1/n)$ but do you know also other examples?

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These are two quite unrelated questions, I think you should split them up and post as two separate ones. –  Herng Yi Mar 12 '13 at 13:19
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2 Answers

up vote 1 down vote accepted

The condition for the two metrics are equivalent that has you exposed is sufficient but not necessary in general metric spaces. Two metrics are equivalent iff induces the same topology, in particular, when we work in $\mathbb{R}^n$ your condition is also necessary.

Note that $d_\infty\leq d_2\leq \sqrt{n}d_\infty$. Then $d_\infty$ and $d_2$ are equivalent.

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We have better $d_2\leq \sqrt{n}d_{\infty}$. –  Sami Ben Romdhane Mar 12 '13 at 13:27
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Regarding your second question:

  • Clearly your example can be generalised to $\Bbb R^n$ with the Euclidean (standard) topology: just choose a point and take the open balls of radius $\frac{1}{n}$ centred on that point. Actually, any sequence of real numbers converging to $0$ will do as radii.

  • Consider $X=\Bbb R$ (a line) with the cofinite topology, i.e. where the open sets are only the complements of finite sets and the empty set. Then the intersection of an infinite number of open sets is the complement of an infinite set, hence it isn't open.

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