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If $X,X'$ have the same $\varepsilon$-packing numbers and $f:X \to X'$ surjective $1$-Lipschitz, then $f$ is an isometry
Let $(X, d)$ be a compact metric space.
We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-covering of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, ...
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2
votes
1
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Are two metric spaces isometric if they have the same $\varepsilon$-covering and $\varepsilon$-packing numbers for all $\varepsilon>0$?
Let $(X, d)$ be a compact metric space.
We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-covering of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, ...
- 109k
2
votes
1
answer
259
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Are two metric spaces isometric if they have the same $\varepsilon$-covering numbers for all $\varepsilon>0$?
Let $(E, d)$ be a metric space. For $\varepsilon>0$, we define two notions of $\varepsilon$-covering number as follows, i.e.,
$N_\varepsilon^o (E)$ is the smallest number of open balls whose radii ...
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