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The diameter $D(M, d) \in \mathbb{R} \cup \{\infty\}$ of a metric space $(M, d)$ is $D(M, d) = \sup\{d(x, y) : x, y \in M\}$ if $M$ is not empty and 0 if $M = \emptyset$. Find a distance $d′$ on $\mathbb{R}$ with the same convergent sequences as the standard distance $d$ but so that $D(\mathbb{R}, d′)$ is finite.

so what I was thinking is to change the domain for d' so that it has a finite diameter ?

mephistolotl
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excalibirr
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2 Answers2

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Put $d'(x,y) = |\arctan(x) - \arctan(y)|$. Then $D(\mathbb R,d') = \pi$. $d'$ is obviously a metric. Moreover, $d'(x,y)\le |x-y|$ since $\arctan'(x) = \frac 1 {1+x^2}\le 1$. Hence, usual Cauchy sequences are also $d'$-Cauchy. The fact that $d'$-Cauchy sequences are usual Cauchy follows from the continuity of $\tan$ on $(-\tfrac\pi 2,\tfrac\pi 2)$.

amsmath
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Every metric is equivalent to a bounded metric, where equivalent means that it induces the same topology. There are many ways to construct an equivalent, bounded metric, but the one I remember is $d'(x,y) = \frac{d(x,y)}{1+d(x,y)}$.

You'll have to show

  1. $d'$ is a metric.
  2. A sequence converges w.r.t. $d'$ iff it converges w.r.t. $d$.
mephistolotl
  • 663