Csdrhrt

Is there a countable metric space that isn't a discrete metric space?

The set of rational numbers $mathbb{Q}$ with the usual metric is a countable, metric space (which is also perfect, i.e. it doesn't have any isolated points). Being perfect can be viewed as being 'very far' from discrete.

Bonus: Consider the reals with their natural metric $(mathbb R; d)$. Let $g$ be $mathrm{Coll}(omega, mathbb R)$-generic over $V$. In the generic extension $V[g]$ we have that $mathbb{R}$ is countable but (by an easy absoluteness argument) $(mathbb{R}, d)$ remains a (non-complete but perfect) metric space in this larger universe.

(This can also be achieved by taking a countable, transitivized Skolem hull $M$ of $V_{omega_1}$ and considering $(mathbb R;d)^M$ -- $M$'s version of the reals and their metric. The advantage of the latter construction is that it takes place purely in $mathrm{ZF}$, without stepping out of the set theoretical universe.)

Just for fun I put the formula "metrizable + ~discrete + countable" into https://topology.jdabbs.com/

and found some fun examples of nondiscrete countable metric spaces.

The first example I hadn't heard of before was the "Countable Fort Space" which is defined in the following way:

Let $X$ be a countable set and $pin X$. Define $Usubseteq X$ to be open provided that $Xsetminus U$ is finite or $pnotin U$.

Likewise there is the evenly spaced integer topology.

Let $X=mathbb{Z}$ equipped with the topology generated by all sets of the form

$$a+kmathbb{Z}={a+knmid ninmathbb{Z}}text{ for all }a,kinmathbb{Z}$$

There are others of course, but those were some fun ones that haven't been mentioned in this post yet.

For a metric space to be discrete we need to have every singleton to be open.

Consider the metric space of $Q$, the rational numbers with the standard metric.

$Q$ is countable but singletons are not open because in every neighborhood of a singleton there are points other the singleton itself so no open interval in included in the singleton.

If $X$ is a metric space and $x in X$, then the following statements are equivalent:

• The metric-induced topology on $X$ is not the discrete topology, with ${ x }$ being non-open in $X$.




• $x$ is an accumulation point in $X$.




• There exists a sequence $x_1, x_2, x_3, dots in X setminus { x } $ such that $x_n to x$.

So probably the "sparsest" way to construct a countable metric space $X$ whose metric-induced topology is not the discrete topology is to choose a convergent sequence $x_n$ in $mathbb R$, with limit $x := lim_{n to infty } x_n$ such that the $x_n neq x$ for all $n$, and define $$X = { x_n : n in mathbb N } cup { x }.$$
For example, $X = { frac 1 n : n in mathbb N } cup { 0 }$ works.

Of course, $X = mathbb Q$ works too, as Stefan explained, as witnessed by the fact that it contains the sequence $x_n = frac 1 n$ and its limit $0$.