# Topological spaces which are not pseudometrizable.

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 $$\tau$$.

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.

• 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? – user 170039 May 31 at 13:22
• @user170039 there probably is (I've seen such things at conferences and immediately forgot), but what would be the point? – Henno Brandsma May 31 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. – user 170039 May 31 at 14:53
• In case you happen to find reference for any such thing, please let me know. It will help me immensely. – user 170039 Jun 2 at 3:27

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 $$\{\varnothing, \{0\}, \{0,1\}\}.$$