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Let $(X, d)$ be a metric space. Are $d$ and $d' = \min (d(x,y), 1)$ strongly equivalent?

From the definition, it is clear that $d'(x,y)= \min (d(x,y), 1)\leq d(x,y). $ Here I get $\beta=1$. I tried to prove the reverse inequality for some $\alpha$, I couldn't. I think these two are not equivalent.

Math geek
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    They may or may not, that depends on $(X, d)$. – Martin R Sep 23 '18 at 07:54
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    okay.thank you. – Math geek Sep 23 '18 at 09:26
  • A common term for strongly equivalent is uniformly equivalent. ...$d$ and $d'$ are always equivalent: They generate the same topology. Example: Let $X=\Bbb R$ and $d(x,y)=|x-y|.$ There is no upper bound for ${d(x,y)/d'(x,y):x\ne y}$. Because $d(0,n)=n$ and $d'(0,n)=1$ for $n\in \Bbb N.$ So in this example, $d,d'$ are equivalemt but not uniformly equivalent. – DanielWainfleet Sep 23 '18 at 13:23

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$d$ and $d_1$ are strongly equivalent $\iff$ $d$ is bounded.

$\implies$: If $d$ and $d_1$ are strongly equivalent then for some $\alpha > 0$ and all $x, y \in X$ $$ d(x, y) \le \alpha d_1(x, y) \le \alpha \, , $$ so that $d$ is bounded.

$\impliedby$: If $d(x, y) \le K$ for some $K > 0$ and all $x, y \in X$, then $$ d_1(x, y) \le d(x, y) \le \max(1, K) d_1(x, y) $$ so that the metrics are strongly equivalent. The last inequality holds because both $$ \begin{aligned} d(x, y) &\le K \cdot 1 \le \max(1, K) \cdot 1 \\ d(x, y) &= 1 \cdot d(x, y) \le \max(1, K) \cdot d(x, y) \end{aligned} $$ are true.

Martin R
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