# Is the difference between metric spaces and ultrametric spaces topological?

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)

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

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 $$A \subset C \subset B$$.