Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
I’ve answered the last question, but I’m not really sure what you want explained about the rest. Those topologies aren’t quite right: everywhere that you have $a$ or $b$, you should have $\{a\}$ or $\{b\}$. –  Brian M. Scott Nov 13 '12 at 1:13

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
@ Brian,thanks Brian for the answer.Also I want to know how can be the trivial topology is Pseudometric and Discrete topology is a metric space?Can you please explain me? –  ccc Nov 13 '12 at 1:13
1  
@ccc: I’ll add those to my answer. –  Brian M. Scott Nov 13 '12 at 1:17
    
Thank you very much Brian! –  ccc Nov 13 '12 at 1:23
    
@ccc: You’re welcome! –  Brian M. Scott Nov 13 '12 at 1:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.