is omega a subset of the reals? or just any set. Are there any other conditions on f?
I've spent a lot of time thinking about this problem, but I'm running out of time and need a skeleton of the process to solve it if someone wants to tackle it...
If such that the set is uncountable.
I've been trying to do this by finding a counterexample to the following sentence
let the set is countable.
If I can find a single counterexample, then the original problem is true. I've looked at unions to see if I can get high enough in the ordinals so that uncountably many things drop to when applying the function, but that would mean is true. Which, according to my professor, it is not.
Thus, I need help.... please.
In this case, is the first uncountable ordinal. f is just a function that maps where the output is less than the input or equal to 0.
For example, if we were using the naturals the following could be possible:
f(0) = 0
f(1) = 0
f(2) = 0 or 1
f(3) = 0 or 1 or 2
f(n) = anything less than n
So, f is not actually defined in the problem, that's what I am trying to do; define f so that it maps an uncountable number of things to . I hope that clarifies things.
I'm pretty sure to prove that statement you have to show that any function that satisfies the criteria will have a a_0 so that the preimage of a_0 is uncountable. I'll think about this and see if I can come up with anything, my intuition says that to show uncountablility you may have to use cantor's trick: Cantor's diagonal argument - Wikipedia, the free encyclopedia this is just my intuition though, I'll think about it.
What about doing this with limit ordinals? Say we choose an and look at the limit of it's inverse image, then we do the same for everything in . So, the next thing we choose above must come from above the highest limit ordinal that anything in mapped to. If we keep going with this, we'll eventually find a that maps to itself, a contradiction. Thus, the assumption that everything in our set is countable is false, making the original statement true.