Example of a metric space first countable but not second countable.

Question

I already had a clue about discrete spaces, and coutability, but with the discrete topology, but I do not have clue with metric spaces that are first countable and not second countable.

Answer 1

The standard example is $\mathbb{R}$ (or any other uncountable set) in the discrete metric, $$d(x,y) = \begin{cases} 1 & x\neq y\\ 0 & x=y \end{cases}$$

which induces the discrete topology on $\mathbb{R}$ (all subsets are open because all singletons $\{x\} =B_d(x,1)$ are open.

A base for this topology must include all singletons, and so is always uncountable. So $(\mathbb{R}, d)$ is not second countable, and all metric spaces are first countable (in this case particularly easy to see, as $\{\{x\}\}$ is a local base at each $x$).

Slightly less boring examples are the post-office metric on $\mathbb{R}^2$: if $\|.\|$ is the usual norm on the plane, this metric is defined as

$$d_p(x,y) = \begin{cases} 0 & x=y\\ \|x\| + \|y\| & x \neq y \end{cases}$$

where the name derives from the fact that we measure all distances via a central "post-office" (the origin, here). In this metric, all $x \neq 0$ are isolated points too: $B_{d_p}(x, \|x\|) = \{x\}$, and again it follows that this metric space is not second countable.

Or use the river metric on the plane (where there are direct paths orthogonal to the "river" $y=0$,and all distances use those paths plus the river:

$$d(x,y) = \begin{cases} |x_2 - y_2| & x_1= y_1\\ |x_2| + |y_2| + |x_1 - y_1| & x_1 \neq y_1 \end{cases} $$

Then all sets $\{x\} \times (0,\infty) \subseteq \mathbb{R}^2$ are open and pairwise disjoint and this also shows that no base for it can be countable.

Another classic example is the radial or French metro metric, a variation on the above themes, see here e.g. Note that the unit circle is uncountable and discrete in this metric.

An example from analysis: $\ell^\infty$ the set of all bounded real sequences in the supremum norm.