# Order Topology vs. the standard one?

• Mar 9th 2011, 01:32 AM
aharonidan
Order Topology vs. the standard one?
on \$\displaystyle R^2\$ we define an order topology (lexicographic order) as follow: \$\displaystyle (x1,y1)>(x2,y2)\$ if \$\displaystyle x1>x2\$ or (\$\displaystyle x1=x2\$ & \$\displaystyle y1>y2\$).
Is this topology equivalent to the standard topology? if not, which topology is finer?(give an example of an open set in one but not in the other)?

thanks :)
• Mar 9th 2011, 02:16 AM
tonio
Quote:

Originally Posted by aharonidan
on \$\displaystyle R^2\$ we define an order topology (lexicographic order) as follow: \$\displaystyle (x1,y1)>(x2,y2)\$ if \$\displaystyle x1>x2\$ or (\$\displaystyle x1=x2\$ & \$\displaystyle y1>y2\$).

You have defined no topology at all: you've defined a (partial) order on \$\displaystyle \mathbb{R}^2\$ .

Now, as far as I know, an order topology is defined on a totally ordered set, so if you meant

this then you first prove the above gives you a total order on the real plane and then look at

the derived (order) topology, and then you can ask yourself whether this toplogy is the

same as the usual Euclidean one.

Tonio

Is this topology equivalent to the standard topology? if not, which topology is finer?(give an example of an open set in one but not in the other)?

thanks :)

.
• Mar 9th 2011, 02:43 AM
aharonidan
I wasn't clear enough. if (X,<) is a partial order set, then all (a,b) a<b U (-infinity,infinity) is a base of a topology on X.
now I need to compare this topology with the Euclidean one.
any help or hint is appreciated