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Exercise

Prove that every metric space $(X,d)$ admits a metric $d'$ equivalent to $d$ which makes $X$ bounded.

EDIT

Definition

Let $X$ be a non-empty set. We say the metrics $d_{1}$ and $d_{2}$ defined over $X$ are equivalent iff for every $x\in X$ and every $r > 0$:

(a) exists $s > 0$ such that $\{y\in X : d_{1}(x,y) < s\} \subset \{y\in X : d_{2}(x,y) < r\}$,

(b) exists $t > 0$ such that $\{y\in X : d_{2}(x,y) < t\} \subset \{y\in X : d_{1}(x,y) < r\}$.

My attempt

Based on the suggestion of @TheoBendit, if we take $d'(x,y) = \min\{d(x,y),1\}$, one has that for every $x\in X$ and every $r > 0$ there exists $s = \min\{r,1\}$ such that
\begin{align*} \{y\in X : d'(x,y) < s\} \subset \{y\in X : d(x,y) < r\}. \end{align*}

But I fail to prove the converse.

Any help is appreciated.

user1234
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    $d' = \min{d, 1}$ should work. – Theo Bendit Jan 31 '22 at 00:28
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    It's equivalent in that they generate the same topology. It's not equivalent in the sense that $c_1 d' \le d \le c_2 d'$ for some $c_1, c_2 > 0$. Indeed, if $d$ is unbounded, then it is impossible to find a bounded $d'$ satisfying these conditions. – Theo Bendit Jan 31 '22 at 00:31
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    As it happens, your metric would work. (Proving this happens to be exercise $1.2.11$ in Kreyszig's Introductory Functional Analysis textbook, hence how I recalled it.) Though it feels like your definition of boundedness might be different? What do you mean by $d' \le d$? And what do you mean by "bounded" in a precise manner? – PrincessEev Jan 31 '22 at 00:36
  • In the following question one other bounded metric is also mentioned: https://math.stackexchange.com/questions/2863127/show-that-for-every-metric-d-the-metrics-d-1d-and-min-1-d-are-equ?rq=1 . With the notation of that question, you can check that $d_1(x,y) < 1$ for all $x,y \in X$. – Kooranifar Jan 31 '22 at 21:15

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