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.

I'm a newcomer in topology, so I have many things chaotic in my minds, so I hope you could help me. In order topology, an basis has structure $(a,b)$, right. This is no problem when considering a topology like R, but, what if the number of elements between a and b is finite, so we can write $$(a,b) = [a_1, b_1]$$ which is not open, right? Can any one explain to me. Thanks

share|improve this question
    
Thanks so much ^_^ –  le duc quang Sep 16 '12 at 16:44

2 Answers 2

up vote 2 down vote accepted

In the case of a finite ordered set $X$, the order topology is discrete. In particular, this implies that for any $a,b\in X$, $[a,b]$ is open (as a union of open sets). It is closed, yes, but it is also open. (Perhaps your point of difficulty is thinking that closed sets cannot also be open - this is not true, since in particular you have observed a counterexample!)

Hint for proving that the order topology on a finite set is discrete: How would you show that singletons are open?

share|improve this answer
1  
I think to prove that order finite topology is discreet, we write ${a} = (a_1, b_1)$, here $a_1, b_1$ are siblings of $a$ in order relation, so ${a}$ is basis element, right? –  le duc quang Sep 16 '12 at 16:18
1  
@le duc quang: Almost right: if $a_1$ is the immediate predecessor of $a$, and $b_1$ is the immediate successor of $a$, then $\{a\}$ is a basic open set. (Not $a$, but the set whose only member is $a$.) –  Brian M. Scott Sep 16 '12 at 20:07

In some ordered sets like $[0,1]$, in order to get a base for the order topology, you need consider too the intervals of the form $(\leftarrow,a)$ and $(b,\rightarrow)$.

share|improve this answer
1  
That gives you a subbase, not a base. –  Michael Greinecker Sep 18 '12 at 15:47
    
In $R$ it is true –  Gustavo Rubiano Sep 18 '12 at 17:11
    
$(a,b)$ is not the union of infinite halflines, but certainly open in the order topology. –  Michael Greinecker Sep 18 '12 at 17:13
    
In $[0,1]$ with the order topology, the all space is not open if we only consider intervals of the form $(a,b)$ and not infinite halflines. –  Gustavo Rubiano Sep 18 '12 at 17:20
1  
There is something wrong with a "topology" in which the ambient space is not open... –  Michael Greinecker Sep 18 '12 at 21:54

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.