Metric topologies on $\mathbb{R}$

Question

I know of three metrics that induce distinct topologies on $\mathbb{R}$: the standard $d(x,y)=|x-y|$, $p$-adic, and discrete $d(x,y)=\boldsymbol{1}[x\neq y]$. Does this exhaust the possibilities? Or are there other metric topologies on $\mathbb{R}$?

Answer 1

There are lots of possible metrics on $\mathbb{R}$. Those you've mentioned is just a tiny, tiny fraction (if it even makes sense).

Consider any metric space $(X,d)$ of cardinality $\mathfrak{c} $. Then any bijection $f:\mathbb{R}\to X$ induces a metric $d'$ on $\mathbb{R}$ by $d'(x,y):=d(f(x),f(y))$. With this $f$ becomes a homeomorphism. Moreover if $(X,d)$ is not homeomorphic to $(X',d')$, then those transfered topologies cannot be homeomorphic as well. And so induced topologies have to be at least distinct.

In particular we can transfer for example the structure of any manifold of positive dimension to $\mathbb{R}$. So you can turn $\mathbb{R}$ into a space of any dimension (even infinite), compact or not, connected or not, and so on, and so on. You can put all the weird CW structures on $\mathbb{R}$. You can make it totally disconnected (e.g. by transfering metric from irrationals). Each of those ideas potentially generates a distinct (even up to homeomorphism) metric on $\mathbb{R}$.

Without additional restrictions, the number of possibilities is just huuuge. More precisely, it is $2^{\mathfrak{c}}$, even up to homeomorphism:

Consider $T(A)=\mathbb{R}^2\sqcup A$ where $A\subseteq\mathbb{R}$. This space is metrizable, is of cardinality $\mathfrak{c}$ (regardless of $A$) and $T(A)$ is homeomorphic to $T(B)$ if and only if $A$ is homeomorphic to $B$ (crucial observation here is that no subset of $\mathbb{R}$ is homeomorphic to $\mathbb{R}^2$). So how many non-homeomorphic subsets of $\mathbb{R}$ are there? Let $X\in\mathcal{P}(\mathbb{R})$. Since $X$ is separable, then every embedding $X\to\mathbb{R}$ is fully determined by values on some countable (possibly finite) subset. So there are at most $2^{\aleph_0}$ such embeddings. This implies that at most $2^{\aleph_0}$ members of $\mathcal{P}(\mathbb{R})$ can be homeomorphic to $X$ (because each comes with a different embedding into $\mathbb{R}$: the inclusion). But $\mathcal{P}(\mathbb{R})$ has $2^{\mathfrak{c}}$ members. And thus, by the arbitrary choice of $X$ we get that there are $2^{\mathfrak{c}}$ homeomorphism classes in $\mathcal{P}(\mathbb{R})$.

Answer 2

Let $B_n$ be the unit ball in $\mathbb{R}^n$. Let $G$ be any connected graph (not necessarily locally finite), and let $N\colon V(G)\to \mathbb{Z}_{\geq 2}$ be a labelling of the vertices.

Construct a metrizable space by taking a copy of $B_{N(v)}$ for each $v$, an copy of $[0,1]$ for each edge, and gluing the edge to two points in the balls at its ends (with the path metric you get something not entirely unlike an $\mathbb{R}$-tree even for vertices with infinite degree). By the theorem on invariance of domain every homeomorphism of such spaces must map edges to edges (thus vertices to vertices) and also preserve the dimensions of the balls.

This easily produces continuum many nonhomeomorphic topologies (simply by taking nonisomorphic countable locally finite graphs). On the other hand there are $2^{2^\aleph}$ many topologies on the continuum, so I would be surprised if there aren't at least $2^\aleph$ metrizable ones.