I am trying to find out whether or not this is true.
Let be an arbitrary topological space, and let have the order topology. If are continuous, and if for a specific , then there exists an open set , containing , such that for all .
If I can show this, then my question in the other thread becomes easier.
So far, my intuition tells me that the best way to proceed would be to use the definition of continuity that says is continuous at a point if for each open set containing , we can find an open set , containing , such that . I haven't gotten much further than that, though.
-Steve
The fact you state is one that I would use in the last steps of this proof; after showing that around each I could form an open such that for all , I would then claim that the set must be open since around each of its members we can form an open set entirely contained within . But this fact does not help work out the intermediate steps in the problem.
What I need help with is showing that the beginning of the above paragraph is actually true: that around each I can form an open such that for all .
Yes. This is a detailed description.
Let defined by h(x)=f(x)-g(x) for all x in X. We know that f and g are continuous on X. Thus, h is continuous on X.
Let . Let V be an open set containing a point y_0 in B and in contained in B. Since h is continuous, we have an open set U containing such that . It follows that an arbitrary point in U, let's say x_1, satisfies that . Thus, for all .
Sorry, this is my studpid mistake introducing addition operation and 0.
I updated the previous post. See here
In the link, is the required open set containing x_0, satisfying f(x) > g(x) for all points x inside the open set .
So, I think your assumption is true.
Hope some other people verify this as well.