Ultrametrizable spaces

properties/P000220.md

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+---
+uid: P000220
+name: Ultrametrizable
+refs:
+  - wikipedia: Ultrametric_space
+    name: Ultrametric space on Wikipedia
+---
+
+There exists an ultrametric inducing the topology of $X$.
+
+A metric $d$ is called an *ultrametric* if it satisfies a property stronger than the triangle inequality, namely:
+
+$\quad d(x, z) \leq \max(d(x, y), d(y, z)).$

theorems/T000770.md

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+---
+uid: T000770
+if:
+  P000220: true
+then:
+  P000053: true
+---
+
+By definition.

theorems/T000771.md

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+---
+uid: T000771
+if:
+  P000220: true
+then:
+  P000146: true
+refs:
+  - doi: 10.48550/arXiv.1306.6086
+    name: Ultraparacompactness and Ultranormality (J. Van Name)
+---
+
+If $d$ is an ultrametric on $X$, then $d(x, A)\leq \max(d(x, y), d(y, A))$. This implies that if $d(x, y) < d(x, A)$, then $d(x, A) = d(y, A)$.
+
+We first show $X$ is {P218}.
+Let $A$ and $B$ be nonempty disjoint closed sets.
+Define $f:X\to[0,1]$ by $f(x) := \frac{d(x, A)}{d(x, A)+d(x, B)}$.
+The set $C = f^{-1}(1/2)$ is closed.
+If $x\in C$, then $r := d(x, A) = d(x, B) > 0$ so that $d(x, y) < r$ implies $d(y, A) = d(x, A) = d(x, B) = d(y, B)$, and therefore $B(x, r)\subseteq C$.
+It follows that $C$ is clopen, so that $A, B$ are contained in the disjoint clopen sets $f^{-1}([0, 1/2))$ and $f^{-1}((1/2, 1])$.
+
+The result then follows since {P53} {P218} spaces are {P146}
+[(Explore)](https://topology.pi-base.org/spaces?q=Metrizable%2BUltranormal%2B%7EUltraparacompact).
+
+See also Theorem 11 in {{doi:10.48550/arXiv.1306.6086}}.

theorems/T000772.md

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+---
+uid: T000772
+if:
+  and:
+  - P000053: true
+  - P000146: true
+then:
+  P000220: true
+---
+
+Let $d$ be a metric on $X$. Define $\mathcal{U}_1 = \{X\}$ and $\mathcal{U}_{n+1}$ to be an open partition which is a refinement of $\{B_d(x, \frac{1}{n})\cap U : x\in X, U\in\mathcal{U}_n\}$. Define $\delta(x, y) = \inf\{1/n : \{x, y\}\subseteq U \text{ for some }U\in\mathscr U_n \}$.
+
+- (1) $\delta(x, y) = 0$ iff $x = y$
+
+  If $x = y$, then there is unique $U_n\in \mathcal{U}_n$ such that $x\in U_n$, and so $\delta(x, y) = 0$. Conversely, if $\delta(x, y) = 0$, then there are $U_n\in \mathcal{U}_n$ such that $\{x, y\}\subseteq U_n$ for all $n$. Then $\text{diam } \{x, y\} \leq \text{diam } U_{n+1} \leq 2/n$ and so $x = y$.
+
+- (2) $\delta(x, y) = \delta(y, x)$
+
+  Clear from the definition.
+
+- (3) $\delta(x, z)\leq \max(\delta(x, y), \delta(y, z))$
+
+  If $\delta(x, z)> \max(\delta(x, y), \delta(y, z))$ then there exists $n$ such that $\{x, y\}\subseteq V$ and $\{y, z\}\subseteq U$ for some $V, U\in \mathcal{U}_n$ and $\{x, z\}$ is not contained in an element of $\mathcal{U}_n$. Since $\mathcal{U}_n$ is a partition and $y\in U\cap V$ it follows that $U = V$ and so $\{x, z\}\subseteq U$, which is a contradiction.
+
+- (4) $\delta$ and $d$ induce the same topology on $X$
+
+  Let $x\in X$ and $r > 0$ be given. Observe that $B_\delta(x, \frac{1}{n}) = U$ where $U\in \mathcal{U}_{n+1}$ is the unique element in the partition containing $x$. Then $B_\delta(x, r) = B_\delta(x, \frac{1}{n})$ or $B_\delta(x, r) = X$ and so $B_\delta(x, r)$ is open, so there is $s > 0$ such that $B_d(x, s)\subseteq B_\delta(x, r)$. And since $\text{diam }U \leq \frac{2}{n+1}$ for $U\in\mathcal{U}_{n+1}$, there is $n$ such that $B_\delta(x, \frac{1}{n})\subseteq B_d(x, r)$. And so the metrics $d, \delta$ are topologically equivalent.
+
+Since $\delta$ is an ultrametric, it follows that $X$ is {P220}.