if f(x0)>g(x0) for certain x0,let u=(f(x0)-g(x0))/3
then there exist v such that for any |x-x0|<v, |f(x)-f(x0)|<u,|g(x)-g(x0)|<u because of the continuuity.
Thus f(x)>g(x) for any |x-x0|<v
So it will be an open set.
I'm sure this is pretty elementary but I am intellectually needy.
I need to show that with both continuous and with in the order topology, the set is closed.
My work so far: I am trying to show that the complement is open. If we denote the given set by , then the complement is . For a particular satisfying we have one of two cases:
(1) is the immediate successor of . In this case, let be defined by , and let be defined by .
(2) There exists some satisfying ; in this case, define as , and define as for some such .
In both of these cases, are open sets (since has the order topology), and since are continuous, we have that and are both open in .
That's the work that I've done so far, with some coaching from a professor. Now what I would like to do is create a collection of sets , and then claim that since is open for each ( is open and is continuous), we can take the union and claim it to be open in . This is where my reasoning gets a little blurry, though. Is this union really the same as ? It seems in a certain sense that it is, since each produces the set of inputs such that the outputs of are greater than the outputs of for that particular . But in another sense it seems obvious that this does NOT compose the set in question, since if we allow to range over all possible values then the union will include a great many values for which in general it is not true that . ALSO, and perhaps more vexingly, the current argument does not make use of the continuity of .
Can anyone help with this? Thanks!
The problem does not state that the functions in question are real-valued, and so we cannot use things like "division by 3" in the answer. If they were both real-valued then the problem would be very easy. But we have no measures, no algebraic structure, no binary operations or any other structure on other than that it is a topological space with the order topology and nothing on other than that it is an arbitrary topological space.
For 1, let ; let and . Again, is an open set containing and is contained in (A choice of is arbitrary in as well ). Thus, is open.
Since is open, we conclude that A is closed.