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
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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?
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 $(b,\rightarrow)$.