שמור ב:
| מחבר ראשי: | |
|---|---|
| פורמט: | Preprint |
| יצא לאור: |
2025
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| נושאים: | |
| גישה מקוונת: | https://arxiv.org/abs/2501.08030 |
| תגים: |
הוספת תג
אין תגיות, היה/י הראשונ/ה לתייג את הרשומה!
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תוכן הענינים:
- Given a metrizable space $Z$, denote by ${\rm PM}(Z)$ the space of continuous bounded pseudometrics on $Z$, and denote by ${\rm AM}(Z)$ the one of continuous bounded admissible metrics on $Z$, the both of which are equipped with the sup-norm $\|\cdot\|$. Let ${\rm Pc}(Z)$ be the subspace of ${\rm AM}(Z)$ satisfying the following: \begin{itemize} \item for every $d \in {\rm Pc}(Z)$, there exists a compact subset $K \subset Z$ such that if $d(x,y) = \|d\|$, then $x, y \in K$. \end{itemize} Moreover, set $${\rm Pp}(Z) = \{d \in {\rm AM}(Z) \mid \text{ there only exists } \{z,w\} \subset Z \text{ such that } d(z,w) = \|d\|\},$$ and let ${\rm M}(Z)$ be ${\rm Pc}(Z)$ or ${\rm Pp}(Z)$. In this paper, we shall prove the Banach-Stone type theorem on spaces of metrics, that is, for metrizable spaces $X$ and $Y$, the following are equivalent: \begin{enumerate} \item $X$ and $Y$ are homeomorphic; \item there exists a surjective isometry $T : {\rm PM}(X) \to {\rm PM}(Y)$ with $T({\rm M}(X)) = {\rm M}(Y)$; \item there exists a surjective isometry $T : {\rm AM}(X) \to {\rm AM}(Y)$ with $T({\rm M}(X)) = {\rm M}(Y)$; \item there exists a surjective isometry $T : {\rm M}(X) \to {\rm M}(Y)$. \end{enumerate} Then for each surjective isometry $T : {\rm PM}(X) \to {\rm PM}(Y)$ with $T({\rm M}(X)) = {\rm M}(Y)$, there is a homeomorphism $ϕ: Y \to X$ such that for any $d \in {\rm PM}(X)$ and for any $x, y \in Y$, $T(d)(x,y) = d(ϕ(x),ϕ(y))$. Except for the case where the cardinality of $X$ or $Y$ is equal to $2$, the homeomorphism $ϕ$ can be chosen uniquely.