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
Order topology on a space where intervals are finite
0
$\begingroup$
general-topology
2 Answers
2
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?
-
1I 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? – 2012-09-16
-
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 $a$.) – 2012-09-16
0
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 $(b,\rightarrow)$.
-
1That gives you a subbase, not a base. – 2012-09-18
-
0In $R$ it is true – 2012-09-18
-
0$(a,b)$ is not the union of infinite halflines, but certainly open in the order topology. – 2012-09-18
-
0In $[0,1]$ with the order topology, the all space is not open if we only consider intervals of the form $(a,b)$ and not infinite halflines. – 2012-09-18
-
1There is something wrong with a "topology" in which the ambient space is not open... – 2012-09-18