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My question is exactly what is written in the title. A metric space is a set $X$ equipped with a function $d: X \times X \to \mathbb{R}_{\geq 0}$ satisfying $d(x, y) = 0 \iff x = y$, $d(x, y) = d(y, x)$, and the triangle inequality: $d(x, y) \leq d(x, z) + d(z, y)$ for all $z$. An ultrametric space is a set $X$ equipped with an ultrametric: a metric $d: X \times X \to \mathbb{R}_{\geq 0}$ satisfying the strong triangle inequality: $d(x, y) \leq \max(d(x, z), d(z, y))$ for all $z$. Ultrametric spaces have some strange properties - for example every open ball is also closed, and every point inside a ball is a center of it (that is, if $y \in B_r[x]$ then $B_r[x] = B_r[y]$).

Metrics induce a topology, so metric spaces and ultrametric spaces are topological spaces. For example it is well-known that the topology of a metric space is Hausdorff, first-countable, and sequentially compact (more generally it is paracompact). I am wondering if/to what extent the difference between metric spaces and ultrametric spaces is topological.

Some ideas:

  • It could be that ultrametric spaces in general are not orderly enough to describe topologically. In this case it may be better to consider the more well-behaved ultranormed spaces vs. normed spaces - for a norm/ultranorm $\|-\|: A \to \mathbb{R}_{\geq 0}$ to induce a metric/ultrametric $d: A \times A \to \mathbb{R}_{\geq 0}$ (by $d(a, b) := \|a - b\|$) we need $A$ to be at minimum an abelian group (norms on vector spaces are much more familiar, but norms on abelian groups are the same thing just without the axiom $\|\lambda \cdot v\| = |\lambda| \|v\|$). That is, normed abelian groups satisfy the axiom $\|a + b\| \leq \|a\| + \|b\|$ while ultranormed abelian groups satisfy the stronger axiom $\|a + b\| \leq \max(\|a\|, \|b\|)$. The reason ultranormed spaces are better-behaved than general ultrametric spaces is because the induced ultrametrics are translation-invariant ($d(a, b) = d(a+c, b+c)$), and also the abelian group structure implies that the strong triangle inequality $\|a + b\| \leq \max(\|a\|, \|b\|)$ becomes an equality if $\|a\| \neq \|b\|$, and so the induced ultrametric satisfies $d(x, y) \leq \max(d(x, z), d(z, y))$ for all $z$ with equality if $d(x, z) \neq d(z, y)$.
  • It could be that the difference between metric spaces and ultrametric spaces is not described by the topological structure but rather by the uniform structure, uniform spaces being a generalisation of metric spaces (and of topological abelian groups) introduced by Weil and Bourbaki. The reason that every metric space is a topological space at a deeper level is because every metric space is a uniform space and every uniform space is a topological space. Like with the topology, every metric space is a Hausdorff uniform space (there is a uniform definition of Hausdorffness).
  • One could even combine these two ideas: every normed/ultranormed abelian group is a uniform abelian group which in turn is a topological abelian group.

(Edit: corrected mistake)

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    $\begingroup$ It is well-known that a topological space is ultrametrizable, iff it is metrizable and strongly zero-dimensional. $\endgroup$
    – Ulli
    Commented Jul 19, 2023 at 11:54
  • $\begingroup$ @Ulli Yes sorry my mistake. $\endgroup$ Commented Jul 19, 2023 at 12:06

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The answer is yes, ultrametric spces can be characterised among metric spaces as the strongly zero-dimensional ones. It is an old result of De Groot.

Definition: The space $X$ is said to be strongly zero-dimensional to mean for every closed set $A$ and open set $B$ with $A \subset B$ there is a clopen set $C$ with $A \subset C \subset B$.

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