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Is the metric $D(x,y)=|1/x-1/y|$ equivalent to the standard metric $d$ on $(0,1]$?

If so,how?

I know that I have to show for $\epsilon >0$ there exists $\delta >0$ such that $B_D (x, \delta) \subset B_d (x, \epsilon)$.

Also,what about the metric $P(x,y)=|x/({1+|x|})-y/({1+|y|})|$? Is it equivalent to the usual metric on $R$?

jimm
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  • It is easy to see that if two metrics are equivalent, then sequences that are Cauchy w.r.t. one must also be Cauchy w.r.t. the other. Now consider $x_n := 1/n$ for $n \in \mathbb{N},$ which is clearly Cauchy under $d,$ but $D(x_n, x_m) = |n-m| \ge 1.$ – stochasticboy321 May 30 '18 at 03:10
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    @stochasticboy321 No, that argument only works if they're uniformly equivalent. Equivalent metrics induce the same topology, not necessarily the same uniformity. – Henno Brandsma May 30 '18 at 04:28

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Yes, they are equivalent because $f: ((0,1],D) \to ([1, \infty),d)$ defined by $f(x) = \frac{1}{x}$ is an isometry:

$$d(f(x), f(y)) = |\frac{1}{x} - \frac{1}{y}| = D(x,y)$$

and $((0,1],d)$ and $([1,\infty),d)$ are homeomorphic, vis that same $f$.

Henno Brandsma
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Two metrics are equivalent iff they have the same convergent sequences with the same limits. In the first case this condition is satisfied so the metrics are equivalent. In the case of $P(x,y)$ note that $P(x_n,x) \to 0$ implies $\frac {x_n} {1+|x_n|} \to \frac x {1+|x|}$. Taking absolute value on both sides we get $\frac {|x_n|} {1+|x_n|} \to \frac {|x|} {1+|x|}$. This implies $|x_n| \to |x|$. Going back to $\frac {x_n} {1+|x_n|} \to \frac x {1+|x|}$ we see that $x_n \to x$. Converse part is straightforward. So $P$ is also equivalen to to the standard metric.

Kavi Rama Murthy
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