point set topology-Metric spaces

Question

Consider the two point set $X=\{a,b\}$ The possible topologies that can be found from X as follows. $$\begin{eqnarray} \tau_1&=&\{X,\emptyset\} &\text{Indiscrete topology} \\ \tau_2&=&\{X,\emptyset,{a}\} \\ \tau_3&=&\{X,\emptyset,{b}\} \\ \tau_4&=&\{X,\emptyset,{a},{b}\} &\text{Discrete topology} \end{eqnarray}$$

It is given that the trivial topology is Pseudometric and Discrete topology is a metric space.Also, $\tau_2$ and $\tau_3$ are known as Sierpinski space. Can you please explain me the above facts?

Further,I know that $\tau_2$ and $\tau_3$ are not $T_1$. But can they be Pseudo metric?

Answer 1

No, $\tau_2$ and $\tau_3$ are not pseudometrizable. Suppose that $d$ is a pseudometric generating $\tau_2$. Since $a\in\{a\}\in\tau_2$, there must be some $r>0$ such that $a\in B_d(a,r)\subseteq\{a\}$, where $$B_d(a,r)=\{x:d(a,x)<r\}$$ is as usual the open ball of radius $r$ centred at $a$. Note that $b\notin B_d(a,r)$, so $d(a,b)\ge r$. A pseudometric is symmetric, so $d(b,a)=d(a,b)\ge r$, and therefore $a\notin B_d(b,r)$. Thus, $b$ has an open neighborhood that does not contain $a$. But this is false: the only open set containing $b$ is $\{a,b\}$. Thus, $\tau_2$ cannot in fact be generated by any pseudometric.

The indiscrete topology on any set $X$ is generated by the pseudometric $d$ such that $d(x,y)=0$ for all $x,y\in X$; I’ll leave it to you to check that this really is a pseudometric.

The discrete topology on any set $X$ is generated by the metric $d$ defined by

$$d(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;;\end{cases}$$

you should not have much trouble verifying that this really is a metric.