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how do I show a topological space X with an order topology is regular. I've shown it is hausdorff already.
In the order topology, a space is considered regular if for every point in the space and every open set containing that point, there exists a smaller open set containing the point that is still contained within the larger open set.
Showing regularity in the order topology is important because it ensures that the space is well-behaved and has certain desirable properties, such as being able to separate points and closed sets.
Yes, regularity in the order topology can also be defined as the intersection of all open sets containing a given point being equal to that point, or as every open set containing a given point being contained within another open set containing that point.
In general, regularity is a stronger condition than Hausdorffness. However, in the order topology, regularity and Hausdorffness are equivalent. This means that a space is regular in the order topology if and only if it is Hausdorff.
No, regularity is not a necessary condition for a space to have the order topology. The order topology can be defined on any totally ordered set, regardless of whether it is regular or not.