How could one geometrically visualize any given metric space $(X,d)$?

Question

Example. Say $X=\mathbb{R}$ and $d(x,y)=\frac{d_0(x,y)}{1+d_0(x,y)}$ where $d_0(x,y)=|x-y|$ is the Euclidean metric.

The visualization of e.g. $\mathbb{R}$ with Eucledian distance $d_0(x,y)=|x-y|$ is clear, because it coincides with the intuitive distance between two points on a real line.

However if one changes the metric to be $d(x,y)=\frac{d_0(x,y)}{1+d_0(x,y)}$ it seems not obvious to me what a visualization could look like. Changing the metric of course changes the open and closed sets; how do these changed open and closed sets look on the real line?

Answer 1

Say $(X,d)$ is a metric space. The open discs in $X$ form a base for the topology induced by $d$.

An open disc $D_\epsilon (x)$ with radius $\epsilon>0$ and midpoint $x\in X$ in a metric space $(X,d)$ is defined as $$D_\epsilon (x):=\{y\in X| d(x,y)<\epsilon\}$$

For $X=\mathbb{R}$ and $d(x,y)=\frac{|x-y|}{1+|x-y|}$ these are some examples of open discs $$D_{1/2}(0)=(-1,1)$$ $$D_{2/3}(0)=(-2,2)$$ $$D_{3/4}(0)=(-3,3)$$

Generally:

$$D_{n/(n+1)}(x)=(x-n,x+n)$$

Saying this means, that you can still visualize the open sets in $X=\mathbb{R}$ induced by above metric as unions of open intervals in $\mathbb{R}$ - the changed metric in your case did not alter the open (and the closed) sets / the metrics induce the same topology.

Answer 2

While visualizations may be helpful I suggest actually not focusing on them since that road is doomed to fail. And I will try to explain why:

Changing the metric of course changes the open and closed sets;

Is that so? Well, it might change open balls but all open sets? No, there are many metrics producing the same topology. And there are even different metrics producing same balls, e.g. for real $c>0$ put $d_c(x,y)=c$ and note that all $d_c$ produce the same topology and even exactly the same balls. These are discrete metrics.

how do these changed open and closed sets look on the real line?

Possibly very insane, impossible to comprehend by a mortal ;)

For example for any $n>0$ there is a metric $d$ on $\mathbb{R}$ making $(\mathbb{R}, d)$ homeomorphic to $\mathbb{R}^n$ with the Euclidean metric. This is a consequence of the fact that both $\mathbb{R}$ and $\mathbb{R}^n$ have the same cardinality.

So can you visualize a $12$ dimensional space? And not only that but pack it into the real line? I dare you to do that.

Or lets get it simplier: visualize a discrete metric $d(x,y)=1$ on $\mathbb{R}$. So you start with drawing points separated from each other by a distance $1$ on the plane, right? Wait a minute, on a plane? But $\mathbb{R}$ is uncountable and here's were our intuition collapses: an uncountable subset of a plane cannot be discrete. How can you visualize it then? Because there surely is some quantity making it different from a countable discrete subset that can be embedded into a plane.

But lets get it even simplier. Forget about metrics, what is the real line anyway? How can you visualize it? I mean its like a very dense line, right? But so is $\mathbb{Q}$? And yet we know $\mathbb{R}$ is a lot "bigger". So visualize me a difference between $\mathbb{Q}$ and $\mathbb{R}$. I remember this one being very tricky for me long time ago. I never truely understood it (in terms of intuitive imagination) to be honest. I simply acknowledged it as "it is how it is".

So how come we don't ask questions about visualizing $\mathbb{R}$ to begin with? It doesn't seem easier then exotic metrics. I think that the answer is very simple: the power of habit. We simply got used to the real line, I mean we hear about it since childhood. And the real line itself became a base block for other visualizations.

So here's my answer: get used to exotic metrics and you'll be fine.