Proving tau is pseudometrizable Let $X = \{a, b\}$. Define a topology on X by
$$\tau = \{X, \emptyset, \{a\}\}.$$
Is $\tau$ pseudometrizable? Why? \
I need help with showing the disk $U(a, \epsilon)=\{a\}$.

There are two options: $d(a,b)=0$ and then $U(a,\varepsilon)=\{a,b\}$ or $d(a,b)=\varepsilon >0$ and then $U(b, \varepsilon) = \{b\}$ (it contains $a$ but not $b$, and this is th eonly subset of $X$ that satisfies that) and $U(a, \varepsilon) = \{a\}$ (ditto, as $b$ is not in it, and $a$ is), and so $X$ is discrete. So the only topologies on $\{a,b\}$ that are pseudometrisable are the indiscrete one, and the discrete one.
Another argument: in a pseudometric space we have the symmetry condition (called $R_0$): $$\forall x,y \in X: x \in \overline{\{y\}} \iff y \in \overline{\{x\}}$$ and this is not satisfied for $a,b$ in your example, as $b \in \overline{\{a\}}$ but not conversely. It's caused by the symmetry of the pseudometric, $d(x,y)=d(y,x)$. As soon as we drop that condition (quasi-pseudometrics) your example can be part of it again.
