Is [a,b] ever an open set in the order topology?

[Zoomingout]

My textbook is indicating to me that sometimes {x \in X : a <= x <= b} is an open set. How can this happen?

My only guess is that if X has a smallest and largest element, called a and b, then sure. Otherwise?

 

[Office_Shredder]

Let's look at the natural numbers for inspiration

[3,5]=(2,6)

[4,9]=(3,10)

etc. Hopefully the generalization becomes clear

 

[Zoomingout]

So that would result from our space being the natural numbers and using the order topology. But why is that interval open? I suppose that for any element x in [3,5], we can find an open set, U, around x such that U \subset X.

For 3 we would choose the open set (2,4)?

 

[Office_Shredder]

But why is that interval open?

I'm confused. The definition of an open set in the order topology is that it's a union of open intervals. What do you mean here?

 

[blue_raver22]

Yes, it is possible for the set [a,b] to be open in the order topology. This can occur when the set X has a smallest and largest element, which we can call a and b. In this case, the set [a,b] would consist of all elements in X that are greater than or equal to a and less than or equal to b. Since there is no element in X that is both greater than b and less than a, the set [a,b] would be open in the order topology.

However, if X does not have a smallest and largest element, then [a,b] would not be open in the order topology. In this case, there would be elements in X that are greater than b or less than a, making [a,b] a closed set in the order topology.

It is important to note that in the order topology, a set can be open, closed, or neither depending on the specific set and the elements in X. Therefore, it is possible for [a,b] to be open in some cases and closed in others. This is why it is important to carefully consider the elements and properties of X when determining the openness of a set in the order topology.