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 $\tau$.
Best Answer
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.