5

Let $(X,\tau)$ be a topological space. Then we know some conditions under which $(X,\tau)$ is metrizable (see for example this and this). It is also clear from these theorems that not every topological space is metrizable.

However, I am wondering whether the same is true for pseudometric spaces also. To be more specific,

Does there exist topological spaces which are not pseudometrazible?

Where we agree to call a topological space $(X,\tau)$ to be pseudometrizable iff there exists a pseudometric $d$ on $X$ such that the topology induced by the psuedometric is $\tau$.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199

2 Answers2

9

A pseudometric space is symmetric (also called $R_0$): if $x \in \overline{\{y\}}$ then $y \in \overline{\{x\}}$ (basically because $d(x,y)=0$ implies $d(y,x)=0$ too, also in pseudometric spaces).

Sierpinski space ($X=\{0,1\}$ with topology $\{\emptyset,\{0\},X\}$) is not symmetric so not pseudometrisable. (Because $1 \in \overline{\{0\}}$ but not the other way around). This is in a way the simplest example, certainly the smallest one.

If $X$ is $T_1$ then $X$ is metrisable iff $X$ is pseudometrisable. (the $T_1$ ensures that $X$ is also $R_0$ and so the non-existence of any points $x,y$ with $x \neq y$ but $d(x,y)=0$. So the pseudometric for $X$ on the right is then a metric.)

So spaces like the cofinite topology on $\mathbb{N}$ is not pseudometrisable, as it's not metrisable (not Hausdorff to start with...) Also, the Sorgenfrey line, the Michael line, double arrow space etc etc.

Henno Brandsma
  • 242,131
  • Does there exists any generalization of the notion of metric, say for the time being, a generalized metric such that every topological space is generalized metrizable? –  May 31 '19 at 13:22
  • 2
    @user170039 there probably is (I've seen such things at conferences and immediately forgot), but what would be the point? – Henno Brandsma May 31 '19 at 14:41
  • Actually I am trying to define the notion of differentiable functions for arbitrary topological space. I can already do so for metric spaces (a sketch of idea is given here) and hope to extend these ideal to general topological spaces. –  May 31 '19 at 14:53
  • In case you happen to find reference for any such thing, please let me know. It will help me immensely. –  Jun 02 '19 at 03:27
4

The Sierpinski space is not pseudometrizable. The Kolmogorov quotient of a pseudometric space is metric. However the Sierpinski space is already $T_0$, but it's not Hausdorff, and thus not metric.

The Sierpinski space is the topological space on two points with the topology $$\{\varnothing, \{0\}, \{0,1\}\}.$$

jgon
  • 28,469