Hello,
I'm not sure I understand correctly... Is f a function ? Then what does it mean for a function to be in ?
For each , let be the set , and let be the discrete topology on . For each of the following subsets of , say whether it is open or closed (or neither or both) in the product topology.
(a)
(b)
(c)
(d)
(e)
Recall that
Yes f is a function. Here (the reason for using is because he is using it to refer to the natural as an ordinal, but whatever that is not important). If it helps you can think of as . We define to be the set of all functions that satisfies .
(I would try doing the problem, it is just that I am not sure how "open" sets are defined here. I know there is a way to define "open" for an infinite Cartesian product but I just cannot find a good explanation anywhere on the internet. And I never learned topology . Perhaps you, Moo, know how "open" is defined for infinite Cartesian products).
ThePerfectHacker & Moo, thank you both for helping me (and trying to help). Here is what I understand about this problem now:
So remember that open sets in the infinite product topology is really just having all but finitely many the whole space and the rest are open. Since the individual factors are discrete, you only need to check that all but finitely many are the whole space.
e.g. in (a) the 10th coordinate has a specific value, but all other coordinates can be whatever, so this is certain open.
However, I am still struggling with this problem. Do you have any ideas on the other parts?
Thanks to Opalg [Opalg], we know this:
Take set (b). Let . If then there exists m such that f(m)=0. Then the set is an open neighbourhood of f contained in B. Therefore B is open.
It's usually more difficult to check when a set is closed. You have to look at its complement and decide whether that is open. Sometimes this is straightforward. For example, the complement of set (a) is the set of all f such that f(10)=1. That is open, so set (a) is closed as well as open.
For a slightly less easy example, look at set (c). Let . If then there exists m such that f(m)=f(m+1)=0. Then is an open neighbourhood of f containing no points of C. Therefore the complement of C is open and so C is closed.
I still do not know how to do parts (d.) and (e.)